88376
If ' \(m_{1}\) ' and ' \(m_{2}\) ', \(\left(m_{1}>m_{2}\right)\) are the slopes of the lines represented by \(5 x^{2}-8 x y+3 y^{2}=0\), then \(\mathbf{m}_{\mathbf{1}}: \mathbf{m}_{\mathbf{2}}\) equals
1 \(5: 1\)
2 \(2: 1\)
3 \(5: 3\)
4 \(3: 2\)
Explanation:
(C): Given, \(5 x^{2}-8 x y+3 y^{2}=0\) [dividing by \(x^{2}\) on both side] \(5-\frac{8 y}{x}+3\left(\frac{y}{x}\right)^{2}=0\) Equation of line passes through origin \((0,0)\) \(\mathrm{y}=\mathrm{mx}\) \(\mathrm{m}=\frac{\mathrm{y}}{\mathrm{x}}\) Putting value of \(y / x\) in \(\mathrm{eq}^{\mathrm{n}}\). (1), we get \(5-8 m+3 m^{2}=0\) \(3 m^{2}-8 m+5=0\) \(m^{2}-5 m-3 m+5=0\) \(\mathrm{m}=5,3 \Rightarrow \mathrm{m}_{1}=5, \mathrm{~m}_{2}=3\) \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{5}{3}=5: 3\)
AP EAMCET-2021-19.08.2021
Co-Ordinate system
88377
Let \(P=(-1,0) \mathrm{O}=(0,0)\) and \(\mathrm{Q}=(3,3 \sqrt{3})\) be three points. Then, the equation of the bisector of \(\angle \mathrm{POQ}\) is :
1 \(y=\sqrt{3} x\)
2 \(\sqrt{3} y=x\)
3 \(y=-\sqrt{3} x\)
4 \(\sqrt{3} y=-x\)
Explanation:
(C) : Let the bisector of \(\angle \mathrm{POQ}\) intersects the line \(\mathrm{PQ}\) at \(\mathrm{M}\) \(\mathrm{OP}=1\) \(\mathrm{OQ}=\sqrt{(3 \sqrt{3})^{2}+3^{2}}\) \(\quad \sqrt{27+9}=6\) Now \(\quad \frac{\mathrm{PM}}{\mathrm{OM}}=\frac{\mathrm{OP}}{\mathrm{OQ}}=\frac{1}{6}\) \(\mathrm{PM}: \mathrm{QM}=1: 6\) By section formula \(\left(\frac{-6+3}{1+6}, \frac{0+3 \sqrt{3}}{1+6}\right) \Rightarrow\left(\frac{-3}{7}, \frac{3 \sqrt{3}}{7}\right)\) equation of \(\mathrm{OM}\) \(y-0=\frac{\frac{3 \sqrt{3}}{7}-0}{-\frac{3}{7}-0}(x-0)\) \(-3 y=3 \sqrt{3} x\) \(y=-\sqrt{3} x\)
AMU-2013
Co-Ordinate system
88378
If the lines \(y=3 x+1\) and \(2 y=x+3\) are equally inclined to the line \(y=m x+4\), then the value of ' \(m\) ' is equal to
1 \(\frac{1 \pm 3 \sqrt{2}}{7}\)
2 \(\frac{-1 \pm 5 \sqrt{2}}{7}\)
3 0
4 \(\frac{1 \pm 5 \sqrt{2}}{7}\)
Explanation:
(D) : Given that line \(y=3 x+1\) and \(2 y=x+3\) are equally inclied to the line \(y=m x+4\) we have to find that value of \(m=\) ? Solve case - (i) \(y=3 x+1, y=\frac{x}{2}+\frac{3}{2}\) \(\mathrm{m}_{1}=3 \mathrm{~m}_{2}=\frac{1}{2}\) We know that \(\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) \(\left|\frac{3-m}{1+3 m}\right|=\left|\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\right|\) \(\frac{3-m}{1+3 m}=\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\) \(|x|=|y| \quad x=y\) \(x=-y\) \(\frac{3-m}{1+3 m}=\frac{1-2 m}{2+m}\) \((3-m)(2+m)=(1+3 m)(1-2 m)\) \(=6+3 m-2 m-m^{2}=1-2 m+3 m-6 m^{2}\) \(=5=\mathrm{m}^{2} 5\) \(\mathrm{~m}^{2}=-1\) \(\text { Case } \frac{3-m}{1+3 m}=\frac{-(1-2 m)}{m+2}\) \(-(3-m)(m+2)=-(1-2 m)+(1+3 m)\) \(6+3 m-m^{2}-2 m=-1-3 m+2 m+6 m^{2}\) \(7 m^{2}-2 m-7=0\) \(\mathrm{~m}=\frac{2 \pm \sqrt{4-4 \times 7|-7|}}{2 \times 3}\) \(=\frac{2 \pm \sqrt{4+(1+49)}}{14}=\frac{2 \pm 2 \times 5 \sqrt{2}}{14}\) \(\mathrm{~m}=\frac{1 \pm 5 \sqrt{2}}{7}\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88379
Suppose \(P\) and \(Q\) lie on \(3 x+4 y-4=0\) and \(5 x-\) \(y-4=0\) respectively. If the midpoint of \(P Q\) is \((1,5)\), then the slope of the line passing through \(P\) and \(Q\) is
1 \(\frac{83}{35}\)
2 \(\frac{63}{35}\)
3 \(\frac{-3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(A) : Given that \(P\) and \(Q\) lie an \(3 x+4 y-4=0\) and \(5 \mathrm{x}-\mathrm{y}-4=0\) mid point of \(\mathrm{PQ}\) is \((1,5)\) we have find that slope of the line passing through \(\mathrm{P}\) and \(\mathrm{Q}\) Solve, \(\mathrm{y}-5=\mathrm{m}(\mathrm{x}-1)\) Substituting \(y=m x+5-m\) in the equation \(5 x-y-4=\) \(05 \mathrm{x}-\mathrm{mx}-5+\mathrm{m}-4=0\) \((5-m) x+m-9=0\) So \(\mathrm{x}=\frac{9-\mathrm{m}}{5-\mathrm{m}}\) and \(\mathrm{y}=\mathrm{m}\left(\frac{9-\mathrm{m}}{5-\mathrm{m}}\right) 5-\mathrm{m}=\frac{25-\mathrm{m}}{5-\mathrm{m}}\) Hence, \(\mathrm{p}=\left(\frac{9-\mathrm{m}}{5-\mathrm{m}}, \frac{25-\mathrm{m}}{5-\mathrm{m}}\right)\) Substituting \(y=m x+5-m\) in the equation \(3 x+4 y-\) \(4=0\) We have \(3 x+4(m x+5-m)-4=0\) \((3+4 m) x+16-4 m=0\) \(\text { So, } x=\frac{4 m-16}{4 m+3} \text { and } y=\frac{m(4 m-16)}{4 m+3}+5-m\) \(\frac{4 m^{2}-16 m-4 m^{2}+17 m+15}{4 m+3}=\frac{m+15}{4 m+3}\) \(\text { So, } Q=\left(\frac{4 m-16}{4 m+3}, \frac{m+15}{4 m+3}\right) \tag{i}\) Since \(\mathrm{m}(15)\) is the mid point of \(\mathrm{PQ}\) we have \(1=\frac{1}{2}\left(\frac{9-m}{5-m}+\frac{4 m-16}{4 m+3}\right) \tag{ii}\) \(\text { and } 5=\frac{1}{2}\left(\frac{25-m}{5-m}+\frac{m+15}{4 m+3}\right) \tag{iii}\) From equation we get \(2(5-m)(4 m+3)=(9-m)(4 m+3)+(5-m)(4 m-16)\) \(2\left(-4 m^{2}+17 m+15\right)=\left(-4 m^{2}+33 m+27\right)+\left(-4 m^{2}+\right.\) \(36 \mathrm{~m}-80)\) \(-8 m^{2}+34 m+30=-8 m^{3}+69 m-53\) \(35 \mathrm{~m}=30+53\) \(\mathrm{m}=\frac{83}{35}\) Putting the value of \(m=83 / 35\) in equation - (i) \(\mathrm{Y}-5=8335(\mathrm{x}-1)\) \(83 \mathrm{x}-35 \mathrm{y}+82=0\) The value of \(m\) obtained from equation (iii) is also equal to \(\mathrm{m}=\frac{83}{35}\)
AP EAMCET-2022-05.07.2022
Co-Ordinate system
88381
If the perpendicular bisector of the line segment joining the points \(P(1,4)\) and \(Q(k, 3)\) has \(y\)-intercept equal to -4 , then the value of \(k\) is
1 \(\sqrt{15}\)
2 -4
3 \(\sqrt{14}\)
4 -2
Explanation:
(B) : The slope of Perpendicular bisector to line PQ \(=-\frac{1}{\text { The slope of the original line PQ }}\) \(=-\frac{1}{\frac{3-4}{\mathrm{k}-1}} \Rightarrow(\mathrm{k}-1)\) The midpoint \(=\left(\frac{\mathrm{k}+1}{2}, \frac{7}{2}\right)\) The equation to the bisector 1 is \(\left(\mathrm{y}-\frac{7}{2}\right)=(\mathrm{k}-1)\left(\mathrm{x}-\frac{\mathrm{k}+1}{2}\right)\) As, \(x=0, y=-4\) satisfies it, we have \(\left(-4-\frac{7}{2}\right)=(\mathrm{k}-1)\left(0-\frac{\mathrm{k}+1}{2}\right)=-\frac{15}{2}=-\frac{\mathrm{k}^{2}-1}{2}\) \(\Rightarrow \mathrm{k}^{2}-1=15\) \(\mathrm{k}^{2}=16\) \(\mathrm{k}= \pm 4\) \(\left(-\infty,-\frac{4}{3}\right) \cup\left(\frac{4}{3}, \infty\right)\)
88376
If ' \(m_{1}\) ' and ' \(m_{2}\) ', \(\left(m_{1}>m_{2}\right)\) are the slopes of the lines represented by \(5 x^{2}-8 x y+3 y^{2}=0\), then \(\mathbf{m}_{\mathbf{1}}: \mathbf{m}_{\mathbf{2}}\) equals
1 \(5: 1\)
2 \(2: 1\)
3 \(5: 3\)
4 \(3: 2\)
Explanation:
(C): Given, \(5 x^{2}-8 x y+3 y^{2}=0\) [dividing by \(x^{2}\) on both side] \(5-\frac{8 y}{x}+3\left(\frac{y}{x}\right)^{2}=0\) Equation of line passes through origin \((0,0)\) \(\mathrm{y}=\mathrm{mx}\) \(\mathrm{m}=\frac{\mathrm{y}}{\mathrm{x}}\) Putting value of \(y / x\) in \(\mathrm{eq}^{\mathrm{n}}\). (1), we get \(5-8 m+3 m^{2}=0\) \(3 m^{2}-8 m+5=0\) \(m^{2}-5 m-3 m+5=0\) \(\mathrm{m}=5,3 \Rightarrow \mathrm{m}_{1}=5, \mathrm{~m}_{2}=3\) \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{5}{3}=5: 3\)
AP EAMCET-2021-19.08.2021
Co-Ordinate system
88377
Let \(P=(-1,0) \mathrm{O}=(0,0)\) and \(\mathrm{Q}=(3,3 \sqrt{3})\) be three points. Then, the equation of the bisector of \(\angle \mathrm{POQ}\) is :
1 \(y=\sqrt{3} x\)
2 \(\sqrt{3} y=x\)
3 \(y=-\sqrt{3} x\)
4 \(\sqrt{3} y=-x\)
Explanation:
(C) : Let the bisector of \(\angle \mathrm{POQ}\) intersects the line \(\mathrm{PQ}\) at \(\mathrm{M}\) \(\mathrm{OP}=1\) \(\mathrm{OQ}=\sqrt{(3 \sqrt{3})^{2}+3^{2}}\) \(\quad \sqrt{27+9}=6\) Now \(\quad \frac{\mathrm{PM}}{\mathrm{OM}}=\frac{\mathrm{OP}}{\mathrm{OQ}}=\frac{1}{6}\) \(\mathrm{PM}: \mathrm{QM}=1: 6\) By section formula \(\left(\frac{-6+3}{1+6}, \frac{0+3 \sqrt{3}}{1+6}\right) \Rightarrow\left(\frac{-3}{7}, \frac{3 \sqrt{3}}{7}\right)\) equation of \(\mathrm{OM}\) \(y-0=\frac{\frac{3 \sqrt{3}}{7}-0}{-\frac{3}{7}-0}(x-0)\) \(-3 y=3 \sqrt{3} x\) \(y=-\sqrt{3} x\)
AMU-2013
Co-Ordinate system
88378
If the lines \(y=3 x+1\) and \(2 y=x+3\) are equally inclined to the line \(y=m x+4\), then the value of ' \(m\) ' is equal to
1 \(\frac{1 \pm 3 \sqrt{2}}{7}\)
2 \(\frac{-1 \pm 5 \sqrt{2}}{7}\)
3 0
4 \(\frac{1 \pm 5 \sqrt{2}}{7}\)
Explanation:
(D) : Given that line \(y=3 x+1\) and \(2 y=x+3\) are equally inclied to the line \(y=m x+4\) we have to find that value of \(m=\) ? Solve case - (i) \(y=3 x+1, y=\frac{x}{2}+\frac{3}{2}\) \(\mathrm{m}_{1}=3 \mathrm{~m}_{2}=\frac{1}{2}\) We know that \(\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) \(\left|\frac{3-m}{1+3 m}\right|=\left|\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\right|\) \(\frac{3-m}{1+3 m}=\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\) \(|x|=|y| \quad x=y\) \(x=-y\) \(\frac{3-m}{1+3 m}=\frac{1-2 m}{2+m}\) \((3-m)(2+m)=(1+3 m)(1-2 m)\) \(=6+3 m-2 m-m^{2}=1-2 m+3 m-6 m^{2}\) \(=5=\mathrm{m}^{2} 5\) \(\mathrm{~m}^{2}=-1\) \(\text { Case } \frac{3-m}{1+3 m}=\frac{-(1-2 m)}{m+2}\) \(-(3-m)(m+2)=-(1-2 m)+(1+3 m)\) \(6+3 m-m^{2}-2 m=-1-3 m+2 m+6 m^{2}\) \(7 m^{2}-2 m-7=0\) \(\mathrm{~m}=\frac{2 \pm \sqrt{4-4 \times 7|-7|}}{2 \times 3}\) \(=\frac{2 \pm \sqrt{4+(1+49)}}{14}=\frac{2 \pm 2 \times 5 \sqrt{2}}{14}\) \(\mathrm{~m}=\frac{1 \pm 5 \sqrt{2}}{7}\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88379
Suppose \(P\) and \(Q\) lie on \(3 x+4 y-4=0\) and \(5 x-\) \(y-4=0\) respectively. If the midpoint of \(P Q\) is \((1,5)\), then the slope of the line passing through \(P\) and \(Q\) is
1 \(\frac{83}{35}\)
2 \(\frac{63}{35}\)
3 \(\frac{-3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(A) : Given that \(P\) and \(Q\) lie an \(3 x+4 y-4=0\) and \(5 \mathrm{x}-\mathrm{y}-4=0\) mid point of \(\mathrm{PQ}\) is \((1,5)\) we have find that slope of the line passing through \(\mathrm{P}\) and \(\mathrm{Q}\) Solve, \(\mathrm{y}-5=\mathrm{m}(\mathrm{x}-1)\) Substituting \(y=m x+5-m\) in the equation \(5 x-y-4=\) \(05 \mathrm{x}-\mathrm{mx}-5+\mathrm{m}-4=0\) \((5-m) x+m-9=0\) So \(\mathrm{x}=\frac{9-\mathrm{m}}{5-\mathrm{m}}\) and \(\mathrm{y}=\mathrm{m}\left(\frac{9-\mathrm{m}}{5-\mathrm{m}}\right) 5-\mathrm{m}=\frac{25-\mathrm{m}}{5-\mathrm{m}}\) Hence, \(\mathrm{p}=\left(\frac{9-\mathrm{m}}{5-\mathrm{m}}, \frac{25-\mathrm{m}}{5-\mathrm{m}}\right)\) Substituting \(y=m x+5-m\) in the equation \(3 x+4 y-\) \(4=0\) We have \(3 x+4(m x+5-m)-4=0\) \((3+4 m) x+16-4 m=0\) \(\text { So, } x=\frac{4 m-16}{4 m+3} \text { and } y=\frac{m(4 m-16)}{4 m+3}+5-m\) \(\frac{4 m^{2}-16 m-4 m^{2}+17 m+15}{4 m+3}=\frac{m+15}{4 m+3}\) \(\text { So, } Q=\left(\frac{4 m-16}{4 m+3}, \frac{m+15}{4 m+3}\right) \tag{i}\) Since \(\mathrm{m}(15)\) is the mid point of \(\mathrm{PQ}\) we have \(1=\frac{1}{2}\left(\frac{9-m}{5-m}+\frac{4 m-16}{4 m+3}\right) \tag{ii}\) \(\text { and } 5=\frac{1}{2}\left(\frac{25-m}{5-m}+\frac{m+15}{4 m+3}\right) \tag{iii}\) From equation we get \(2(5-m)(4 m+3)=(9-m)(4 m+3)+(5-m)(4 m-16)\) \(2\left(-4 m^{2}+17 m+15\right)=\left(-4 m^{2}+33 m+27\right)+\left(-4 m^{2}+\right.\) \(36 \mathrm{~m}-80)\) \(-8 m^{2}+34 m+30=-8 m^{3}+69 m-53\) \(35 \mathrm{~m}=30+53\) \(\mathrm{m}=\frac{83}{35}\) Putting the value of \(m=83 / 35\) in equation - (i) \(\mathrm{Y}-5=8335(\mathrm{x}-1)\) \(83 \mathrm{x}-35 \mathrm{y}+82=0\) The value of \(m\) obtained from equation (iii) is also equal to \(\mathrm{m}=\frac{83}{35}\)
AP EAMCET-2022-05.07.2022
Co-Ordinate system
88381
If the perpendicular bisector of the line segment joining the points \(P(1,4)\) and \(Q(k, 3)\) has \(y\)-intercept equal to -4 , then the value of \(k\) is
1 \(\sqrt{15}\)
2 -4
3 \(\sqrt{14}\)
4 -2
Explanation:
(B) : The slope of Perpendicular bisector to line PQ \(=-\frac{1}{\text { The slope of the original line PQ }}\) \(=-\frac{1}{\frac{3-4}{\mathrm{k}-1}} \Rightarrow(\mathrm{k}-1)\) The midpoint \(=\left(\frac{\mathrm{k}+1}{2}, \frac{7}{2}\right)\) The equation to the bisector 1 is \(\left(\mathrm{y}-\frac{7}{2}\right)=(\mathrm{k}-1)\left(\mathrm{x}-\frac{\mathrm{k}+1}{2}\right)\) As, \(x=0, y=-4\) satisfies it, we have \(\left(-4-\frac{7}{2}\right)=(\mathrm{k}-1)\left(0-\frac{\mathrm{k}+1}{2}\right)=-\frac{15}{2}=-\frac{\mathrm{k}^{2}-1}{2}\) \(\Rightarrow \mathrm{k}^{2}-1=15\) \(\mathrm{k}^{2}=16\) \(\mathrm{k}= \pm 4\) \(\left(-\infty,-\frac{4}{3}\right) \cup\left(\frac{4}{3}, \infty\right)\)
88376
If ' \(m_{1}\) ' and ' \(m_{2}\) ', \(\left(m_{1}>m_{2}\right)\) are the slopes of the lines represented by \(5 x^{2}-8 x y+3 y^{2}=0\), then \(\mathbf{m}_{\mathbf{1}}: \mathbf{m}_{\mathbf{2}}\) equals
1 \(5: 1\)
2 \(2: 1\)
3 \(5: 3\)
4 \(3: 2\)
Explanation:
(C): Given, \(5 x^{2}-8 x y+3 y^{2}=0\) [dividing by \(x^{2}\) on both side] \(5-\frac{8 y}{x}+3\left(\frac{y}{x}\right)^{2}=0\) Equation of line passes through origin \((0,0)\) \(\mathrm{y}=\mathrm{mx}\) \(\mathrm{m}=\frac{\mathrm{y}}{\mathrm{x}}\) Putting value of \(y / x\) in \(\mathrm{eq}^{\mathrm{n}}\). (1), we get \(5-8 m+3 m^{2}=0\) \(3 m^{2}-8 m+5=0\) \(m^{2}-5 m-3 m+5=0\) \(\mathrm{m}=5,3 \Rightarrow \mathrm{m}_{1}=5, \mathrm{~m}_{2}=3\) \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{5}{3}=5: 3\)
AP EAMCET-2021-19.08.2021
Co-Ordinate system
88377
Let \(P=(-1,0) \mathrm{O}=(0,0)\) and \(\mathrm{Q}=(3,3 \sqrt{3})\) be three points. Then, the equation of the bisector of \(\angle \mathrm{POQ}\) is :
1 \(y=\sqrt{3} x\)
2 \(\sqrt{3} y=x\)
3 \(y=-\sqrt{3} x\)
4 \(\sqrt{3} y=-x\)
Explanation:
(C) : Let the bisector of \(\angle \mathrm{POQ}\) intersects the line \(\mathrm{PQ}\) at \(\mathrm{M}\) \(\mathrm{OP}=1\) \(\mathrm{OQ}=\sqrt{(3 \sqrt{3})^{2}+3^{2}}\) \(\quad \sqrt{27+9}=6\) Now \(\quad \frac{\mathrm{PM}}{\mathrm{OM}}=\frac{\mathrm{OP}}{\mathrm{OQ}}=\frac{1}{6}\) \(\mathrm{PM}: \mathrm{QM}=1: 6\) By section formula \(\left(\frac{-6+3}{1+6}, \frac{0+3 \sqrt{3}}{1+6}\right) \Rightarrow\left(\frac{-3}{7}, \frac{3 \sqrt{3}}{7}\right)\) equation of \(\mathrm{OM}\) \(y-0=\frac{\frac{3 \sqrt{3}}{7}-0}{-\frac{3}{7}-0}(x-0)\) \(-3 y=3 \sqrt{3} x\) \(y=-\sqrt{3} x\)
AMU-2013
Co-Ordinate system
88378
If the lines \(y=3 x+1\) and \(2 y=x+3\) are equally inclined to the line \(y=m x+4\), then the value of ' \(m\) ' is equal to
1 \(\frac{1 \pm 3 \sqrt{2}}{7}\)
2 \(\frac{-1 \pm 5 \sqrt{2}}{7}\)
3 0
4 \(\frac{1 \pm 5 \sqrt{2}}{7}\)
Explanation:
(D) : Given that line \(y=3 x+1\) and \(2 y=x+3\) are equally inclied to the line \(y=m x+4\) we have to find that value of \(m=\) ? Solve case - (i) \(y=3 x+1, y=\frac{x}{2}+\frac{3}{2}\) \(\mathrm{m}_{1}=3 \mathrm{~m}_{2}=\frac{1}{2}\) We know that \(\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) \(\left|\frac{3-m}{1+3 m}\right|=\left|\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\right|\) \(\frac{3-m}{1+3 m}=\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\) \(|x|=|y| \quad x=y\) \(x=-y\) \(\frac{3-m}{1+3 m}=\frac{1-2 m}{2+m}\) \((3-m)(2+m)=(1+3 m)(1-2 m)\) \(=6+3 m-2 m-m^{2}=1-2 m+3 m-6 m^{2}\) \(=5=\mathrm{m}^{2} 5\) \(\mathrm{~m}^{2}=-1\) \(\text { Case } \frac{3-m}{1+3 m}=\frac{-(1-2 m)}{m+2}\) \(-(3-m)(m+2)=-(1-2 m)+(1+3 m)\) \(6+3 m-m^{2}-2 m=-1-3 m+2 m+6 m^{2}\) \(7 m^{2}-2 m-7=0\) \(\mathrm{~m}=\frac{2 \pm \sqrt{4-4 \times 7|-7|}}{2 \times 3}\) \(=\frac{2 \pm \sqrt{4+(1+49)}}{14}=\frac{2 \pm 2 \times 5 \sqrt{2}}{14}\) \(\mathrm{~m}=\frac{1 \pm 5 \sqrt{2}}{7}\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88379
Suppose \(P\) and \(Q\) lie on \(3 x+4 y-4=0\) and \(5 x-\) \(y-4=0\) respectively. If the midpoint of \(P Q\) is \((1,5)\), then the slope of the line passing through \(P\) and \(Q\) is
1 \(\frac{83}{35}\)
2 \(\frac{63}{35}\)
3 \(\frac{-3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(A) : Given that \(P\) and \(Q\) lie an \(3 x+4 y-4=0\) and \(5 \mathrm{x}-\mathrm{y}-4=0\) mid point of \(\mathrm{PQ}\) is \((1,5)\) we have find that slope of the line passing through \(\mathrm{P}\) and \(\mathrm{Q}\) Solve, \(\mathrm{y}-5=\mathrm{m}(\mathrm{x}-1)\) Substituting \(y=m x+5-m\) in the equation \(5 x-y-4=\) \(05 \mathrm{x}-\mathrm{mx}-5+\mathrm{m}-4=0\) \((5-m) x+m-9=0\) So \(\mathrm{x}=\frac{9-\mathrm{m}}{5-\mathrm{m}}\) and \(\mathrm{y}=\mathrm{m}\left(\frac{9-\mathrm{m}}{5-\mathrm{m}}\right) 5-\mathrm{m}=\frac{25-\mathrm{m}}{5-\mathrm{m}}\) Hence, \(\mathrm{p}=\left(\frac{9-\mathrm{m}}{5-\mathrm{m}}, \frac{25-\mathrm{m}}{5-\mathrm{m}}\right)\) Substituting \(y=m x+5-m\) in the equation \(3 x+4 y-\) \(4=0\) We have \(3 x+4(m x+5-m)-4=0\) \((3+4 m) x+16-4 m=0\) \(\text { So, } x=\frac{4 m-16}{4 m+3} \text { and } y=\frac{m(4 m-16)}{4 m+3}+5-m\) \(\frac{4 m^{2}-16 m-4 m^{2}+17 m+15}{4 m+3}=\frac{m+15}{4 m+3}\) \(\text { So, } Q=\left(\frac{4 m-16}{4 m+3}, \frac{m+15}{4 m+3}\right) \tag{i}\) Since \(\mathrm{m}(15)\) is the mid point of \(\mathrm{PQ}\) we have \(1=\frac{1}{2}\left(\frac{9-m}{5-m}+\frac{4 m-16}{4 m+3}\right) \tag{ii}\) \(\text { and } 5=\frac{1}{2}\left(\frac{25-m}{5-m}+\frac{m+15}{4 m+3}\right) \tag{iii}\) From equation we get \(2(5-m)(4 m+3)=(9-m)(4 m+3)+(5-m)(4 m-16)\) \(2\left(-4 m^{2}+17 m+15\right)=\left(-4 m^{2}+33 m+27\right)+\left(-4 m^{2}+\right.\) \(36 \mathrm{~m}-80)\) \(-8 m^{2}+34 m+30=-8 m^{3}+69 m-53\) \(35 \mathrm{~m}=30+53\) \(\mathrm{m}=\frac{83}{35}\) Putting the value of \(m=83 / 35\) in equation - (i) \(\mathrm{Y}-5=8335(\mathrm{x}-1)\) \(83 \mathrm{x}-35 \mathrm{y}+82=0\) The value of \(m\) obtained from equation (iii) is also equal to \(\mathrm{m}=\frac{83}{35}\)
AP EAMCET-2022-05.07.2022
Co-Ordinate system
88381
If the perpendicular bisector of the line segment joining the points \(P(1,4)\) and \(Q(k, 3)\) has \(y\)-intercept equal to -4 , then the value of \(k\) is
1 \(\sqrt{15}\)
2 -4
3 \(\sqrt{14}\)
4 -2
Explanation:
(B) : The slope of Perpendicular bisector to line PQ \(=-\frac{1}{\text { The slope of the original line PQ }}\) \(=-\frac{1}{\frac{3-4}{\mathrm{k}-1}} \Rightarrow(\mathrm{k}-1)\) The midpoint \(=\left(\frac{\mathrm{k}+1}{2}, \frac{7}{2}\right)\) The equation to the bisector 1 is \(\left(\mathrm{y}-\frac{7}{2}\right)=(\mathrm{k}-1)\left(\mathrm{x}-\frac{\mathrm{k}+1}{2}\right)\) As, \(x=0, y=-4\) satisfies it, we have \(\left(-4-\frac{7}{2}\right)=(\mathrm{k}-1)\left(0-\frac{\mathrm{k}+1}{2}\right)=-\frac{15}{2}=-\frac{\mathrm{k}^{2}-1}{2}\) \(\Rightarrow \mathrm{k}^{2}-1=15\) \(\mathrm{k}^{2}=16\) \(\mathrm{k}= \pm 4\) \(\left(-\infty,-\frac{4}{3}\right) \cup\left(\frac{4}{3}, \infty\right)\)
88376
If ' \(m_{1}\) ' and ' \(m_{2}\) ', \(\left(m_{1}>m_{2}\right)\) are the slopes of the lines represented by \(5 x^{2}-8 x y+3 y^{2}=0\), then \(\mathbf{m}_{\mathbf{1}}: \mathbf{m}_{\mathbf{2}}\) equals
1 \(5: 1\)
2 \(2: 1\)
3 \(5: 3\)
4 \(3: 2\)
Explanation:
(C): Given, \(5 x^{2}-8 x y+3 y^{2}=0\) [dividing by \(x^{2}\) on both side] \(5-\frac{8 y}{x}+3\left(\frac{y}{x}\right)^{2}=0\) Equation of line passes through origin \((0,0)\) \(\mathrm{y}=\mathrm{mx}\) \(\mathrm{m}=\frac{\mathrm{y}}{\mathrm{x}}\) Putting value of \(y / x\) in \(\mathrm{eq}^{\mathrm{n}}\). (1), we get \(5-8 m+3 m^{2}=0\) \(3 m^{2}-8 m+5=0\) \(m^{2}-5 m-3 m+5=0\) \(\mathrm{m}=5,3 \Rightarrow \mathrm{m}_{1}=5, \mathrm{~m}_{2}=3\) \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{5}{3}=5: 3\)
AP EAMCET-2021-19.08.2021
Co-Ordinate system
88377
Let \(P=(-1,0) \mathrm{O}=(0,0)\) and \(\mathrm{Q}=(3,3 \sqrt{3})\) be three points. Then, the equation of the bisector of \(\angle \mathrm{POQ}\) is :
1 \(y=\sqrt{3} x\)
2 \(\sqrt{3} y=x\)
3 \(y=-\sqrt{3} x\)
4 \(\sqrt{3} y=-x\)
Explanation:
(C) : Let the bisector of \(\angle \mathrm{POQ}\) intersects the line \(\mathrm{PQ}\) at \(\mathrm{M}\) \(\mathrm{OP}=1\) \(\mathrm{OQ}=\sqrt{(3 \sqrt{3})^{2}+3^{2}}\) \(\quad \sqrt{27+9}=6\) Now \(\quad \frac{\mathrm{PM}}{\mathrm{OM}}=\frac{\mathrm{OP}}{\mathrm{OQ}}=\frac{1}{6}\) \(\mathrm{PM}: \mathrm{QM}=1: 6\) By section formula \(\left(\frac{-6+3}{1+6}, \frac{0+3 \sqrt{3}}{1+6}\right) \Rightarrow\left(\frac{-3}{7}, \frac{3 \sqrt{3}}{7}\right)\) equation of \(\mathrm{OM}\) \(y-0=\frac{\frac{3 \sqrt{3}}{7}-0}{-\frac{3}{7}-0}(x-0)\) \(-3 y=3 \sqrt{3} x\) \(y=-\sqrt{3} x\)
AMU-2013
Co-Ordinate system
88378
If the lines \(y=3 x+1\) and \(2 y=x+3\) are equally inclined to the line \(y=m x+4\), then the value of ' \(m\) ' is equal to
1 \(\frac{1 \pm 3 \sqrt{2}}{7}\)
2 \(\frac{-1 \pm 5 \sqrt{2}}{7}\)
3 0
4 \(\frac{1 \pm 5 \sqrt{2}}{7}\)
Explanation:
(D) : Given that line \(y=3 x+1\) and \(2 y=x+3\) are equally inclied to the line \(y=m x+4\) we have to find that value of \(m=\) ? Solve case - (i) \(y=3 x+1, y=\frac{x}{2}+\frac{3}{2}\) \(\mathrm{m}_{1}=3 \mathrm{~m}_{2}=\frac{1}{2}\) We know that \(\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) \(\left|\frac{3-m}{1+3 m}\right|=\left|\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\right|\) \(\frac{3-m}{1+3 m}=\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\) \(|x|=|y| \quad x=y\) \(x=-y\) \(\frac{3-m}{1+3 m}=\frac{1-2 m}{2+m}\) \((3-m)(2+m)=(1+3 m)(1-2 m)\) \(=6+3 m-2 m-m^{2}=1-2 m+3 m-6 m^{2}\) \(=5=\mathrm{m}^{2} 5\) \(\mathrm{~m}^{2}=-1\) \(\text { Case } \frac{3-m}{1+3 m}=\frac{-(1-2 m)}{m+2}\) \(-(3-m)(m+2)=-(1-2 m)+(1+3 m)\) \(6+3 m-m^{2}-2 m=-1-3 m+2 m+6 m^{2}\) \(7 m^{2}-2 m-7=0\) \(\mathrm{~m}=\frac{2 \pm \sqrt{4-4 \times 7|-7|}}{2 \times 3}\) \(=\frac{2 \pm \sqrt{4+(1+49)}}{14}=\frac{2 \pm 2 \times 5 \sqrt{2}}{14}\) \(\mathrm{~m}=\frac{1 \pm 5 \sqrt{2}}{7}\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88379
Suppose \(P\) and \(Q\) lie on \(3 x+4 y-4=0\) and \(5 x-\) \(y-4=0\) respectively. If the midpoint of \(P Q\) is \((1,5)\), then the slope of the line passing through \(P\) and \(Q\) is
1 \(\frac{83}{35}\)
2 \(\frac{63}{35}\)
3 \(\frac{-3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(A) : Given that \(P\) and \(Q\) lie an \(3 x+4 y-4=0\) and \(5 \mathrm{x}-\mathrm{y}-4=0\) mid point of \(\mathrm{PQ}\) is \((1,5)\) we have find that slope of the line passing through \(\mathrm{P}\) and \(\mathrm{Q}\) Solve, \(\mathrm{y}-5=\mathrm{m}(\mathrm{x}-1)\) Substituting \(y=m x+5-m\) in the equation \(5 x-y-4=\) \(05 \mathrm{x}-\mathrm{mx}-5+\mathrm{m}-4=0\) \((5-m) x+m-9=0\) So \(\mathrm{x}=\frac{9-\mathrm{m}}{5-\mathrm{m}}\) and \(\mathrm{y}=\mathrm{m}\left(\frac{9-\mathrm{m}}{5-\mathrm{m}}\right) 5-\mathrm{m}=\frac{25-\mathrm{m}}{5-\mathrm{m}}\) Hence, \(\mathrm{p}=\left(\frac{9-\mathrm{m}}{5-\mathrm{m}}, \frac{25-\mathrm{m}}{5-\mathrm{m}}\right)\) Substituting \(y=m x+5-m\) in the equation \(3 x+4 y-\) \(4=0\) We have \(3 x+4(m x+5-m)-4=0\) \((3+4 m) x+16-4 m=0\) \(\text { So, } x=\frac{4 m-16}{4 m+3} \text { and } y=\frac{m(4 m-16)}{4 m+3}+5-m\) \(\frac{4 m^{2}-16 m-4 m^{2}+17 m+15}{4 m+3}=\frac{m+15}{4 m+3}\) \(\text { So, } Q=\left(\frac{4 m-16}{4 m+3}, \frac{m+15}{4 m+3}\right) \tag{i}\) Since \(\mathrm{m}(15)\) is the mid point of \(\mathrm{PQ}\) we have \(1=\frac{1}{2}\left(\frac{9-m}{5-m}+\frac{4 m-16}{4 m+3}\right) \tag{ii}\) \(\text { and } 5=\frac{1}{2}\left(\frac{25-m}{5-m}+\frac{m+15}{4 m+3}\right) \tag{iii}\) From equation we get \(2(5-m)(4 m+3)=(9-m)(4 m+3)+(5-m)(4 m-16)\) \(2\left(-4 m^{2}+17 m+15\right)=\left(-4 m^{2}+33 m+27\right)+\left(-4 m^{2}+\right.\) \(36 \mathrm{~m}-80)\) \(-8 m^{2}+34 m+30=-8 m^{3}+69 m-53\) \(35 \mathrm{~m}=30+53\) \(\mathrm{m}=\frac{83}{35}\) Putting the value of \(m=83 / 35\) in equation - (i) \(\mathrm{Y}-5=8335(\mathrm{x}-1)\) \(83 \mathrm{x}-35 \mathrm{y}+82=0\) The value of \(m\) obtained from equation (iii) is also equal to \(\mathrm{m}=\frac{83}{35}\)
AP EAMCET-2022-05.07.2022
Co-Ordinate system
88381
If the perpendicular bisector of the line segment joining the points \(P(1,4)\) and \(Q(k, 3)\) has \(y\)-intercept equal to -4 , then the value of \(k\) is
1 \(\sqrt{15}\)
2 -4
3 \(\sqrt{14}\)
4 -2
Explanation:
(B) : The slope of Perpendicular bisector to line PQ \(=-\frac{1}{\text { The slope of the original line PQ }}\) \(=-\frac{1}{\frac{3-4}{\mathrm{k}-1}} \Rightarrow(\mathrm{k}-1)\) The midpoint \(=\left(\frac{\mathrm{k}+1}{2}, \frac{7}{2}\right)\) The equation to the bisector 1 is \(\left(\mathrm{y}-\frac{7}{2}\right)=(\mathrm{k}-1)\left(\mathrm{x}-\frac{\mathrm{k}+1}{2}\right)\) As, \(x=0, y=-4\) satisfies it, we have \(\left(-4-\frac{7}{2}\right)=(\mathrm{k}-1)\left(0-\frac{\mathrm{k}+1}{2}\right)=-\frac{15}{2}=-\frac{\mathrm{k}^{2}-1}{2}\) \(\Rightarrow \mathrm{k}^{2}-1=15\) \(\mathrm{k}^{2}=16\) \(\mathrm{k}= \pm 4\) \(\left(-\infty,-\frac{4}{3}\right) \cup\left(\frac{4}{3}, \infty\right)\)
88376
If ' \(m_{1}\) ' and ' \(m_{2}\) ', \(\left(m_{1}>m_{2}\right)\) are the slopes of the lines represented by \(5 x^{2}-8 x y+3 y^{2}=0\), then \(\mathbf{m}_{\mathbf{1}}: \mathbf{m}_{\mathbf{2}}\) equals
1 \(5: 1\)
2 \(2: 1\)
3 \(5: 3\)
4 \(3: 2\)
Explanation:
(C): Given, \(5 x^{2}-8 x y+3 y^{2}=0\) [dividing by \(x^{2}\) on both side] \(5-\frac{8 y}{x}+3\left(\frac{y}{x}\right)^{2}=0\) Equation of line passes through origin \((0,0)\) \(\mathrm{y}=\mathrm{mx}\) \(\mathrm{m}=\frac{\mathrm{y}}{\mathrm{x}}\) Putting value of \(y / x\) in \(\mathrm{eq}^{\mathrm{n}}\). (1), we get \(5-8 m+3 m^{2}=0\) \(3 m^{2}-8 m+5=0\) \(m^{2}-5 m-3 m+5=0\) \(\mathrm{m}=5,3 \Rightarrow \mathrm{m}_{1}=5, \mathrm{~m}_{2}=3\) \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{5}{3}=5: 3\)
AP EAMCET-2021-19.08.2021
Co-Ordinate system
88377
Let \(P=(-1,0) \mathrm{O}=(0,0)\) and \(\mathrm{Q}=(3,3 \sqrt{3})\) be three points. Then, the equation of the bisector of \(\angle \mathrm{POQ}\) is :
1 \(y=\sqrt{3} x\)
2 \(\sqrt{3} y=x\)
3 \(y=-\sqrt{3} x\)
4 \(\sqrt{3} y=-x\)
Explanation:
(C) : Let the bisector of \(\angle \mathrm{POQ}\) intersects the line \(\mathrm{PQ}\) at \(\mathrm{M}\) \(\mathrm{OP}=1\) \(\mathrm{OQ}=\sqrt{(3 \sqrt{3})^{2}+3^{2}}\) \(\quad \sqrt{27+9}=6\) Now \(\quad \frac{\mathrm{PM}}{\mathrm{OM}}=\frac{\mathrm{OP}}{\mathrm{OQ}}=\frac{1}{6}\) \(\mathrm{PM}: \mathrm{QM}=1: 6\) By section formula \(\left(\frac{-6+3}{1+6}, \frac{0+3 \sqrt{3}}{1+6}\right) \Rightarrow\left(\frac{-3}{7}, \frac{3 \sqrt{3}}{7}\right)\) equation of \(\mathrm{OM}\) \(y-0=\frac{\frac{3 \sqrt{3}}{7}-0}{-\frac{3}{7}-0}(x-0)\) \(-3 y=3 \sqrt{3} x\) \(y=-\sqrt{3} x\)
AMU-2013
Co-Ordinate system
88378
If the lines \(y=3 x+1\) and \(2 y=x+3\) are equally inclined to the line \(y=m x+4\), then the value of ' \(m\) ' is equal to
1 \(\frac{1 \pm 3 \sqrt{2}}{7}\)
2 \(\frac{-1 \pm 5 \sqrt{2}}{7}\)
3 0
4 \(\frac{1 \pm 5 \sqrt{2}}{7}\)
Explanation:
(D) : Given that line \(y=3 x+1\) and \(2 y=x+3\) are equally inclied to the line \(y=m x+4\) we have to find that value of \(m=\) ? Solve case - (i) \(y=3 x+1, y=\frac{x}{2}+\frac{3}{2}\) \(\mathrm{m}_{1}=3 \mathrm{~m}_{2}=\frac{1}{2}\) We know that \(\tan \theta=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\) \(\left|\frac{3-m}{1+3 m}\right|=\left|\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\right|\) \(\frac{3-m}{1+3 m}=\frac{\frac{1}{2}-m}{1+\frac{m}{2}}\) \(|x|=|y| \quad x=y\) \(x=-y\) \(\frac{3-m}{1+3 m}=\frac{1-2 m}{2+m}\) \((3-m)(2+m)=(1+3 m)(1-2 m)\) \(=6+3 m-2 m-m^{2}=1-2 m+3 m-6 m^{2}\) \(=5=\mathrm{m}^{2} 5\) \(\mathrm{~m}^{2}=-1\) \(\text { Case } \frac{3-m}{1+3 m}=\frac{-(1-2 m)}{m+2}\) \(-(3-m)(m+2)=-(1-2 m)+(1+3 m)\) \(6+3 m-m^{2}-2 m=-1-3 m+2 m+6 m^{2}\) \(7 m^{2}-2 m-7=0\) \(\mathrm{~m}=\frac{2 \pm \sqrt{4-4 \times 7|-7|}}{2 \times 3}\) \(=\frac{2 \pm \sqrt{4+(1+49)}}{14}=\frac{2 \pm 2 \times 5 \sqrt{2}}{14}\) \(\mathrm{~m}=\frac{1 \pm 5 \sqrt{2}}{7}\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88379
Suppose \(P\) and \(Q\) lie on \(3 x+4 y-4=0\) and \(5 x-\) \(y-4=0\) respectively. If the midpoint of \(P Q\) is \((1,5)\), then the slope of the line passing through \(P\) and \(Q\) is
1 \(\frac{83}{35}\)
2 \(\frac{63}{35}\)
3 \(\frac{-3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(A) : Given that \(P\) and \(Q\) lie an \(3 x+4 y-4=0\) and \(5 \mathrm{x}-\mathrm{y}-4=0\) mid point of \(\mathrm{PQ}\) is \((1,5)\) we have find that slope of the line passing through \(\mathrm{P}\) and \(\mathrm{Q}\) Solve, \(\mathrm{y}-5=\mathrm{m}(\mathrm{x}-1)\) Substituting \(y=m x+5-m\) in the equation \(5 x-y-4=\) \(05 \mathrm{x}-\mathrm{mx}-5+\mathrm{m}-4=0\) \((5-m) x+m-9=0\) So \(\mathrm{x}=\frac{9-\mathrm{m}}{5-\mathrm{m}}\) and \(\mathrm{y}=\mathrm{m}\left(\frac{9-\mathrm{m}}{5-\mathrm{m}}\right) 5-\mathrm{m}=\frac{25-\mathrm{m}}{5-\mathrm{m}}\) Hence, \(\mathrm{p}=\left(\frac{9-\mathrm{m}}{5-\mathrm{m}}, \frac{25-\mathrm{m}}{5-\mathrm{m}}\right)\) Substituting \(y=m x+5-m\) in the equation \(3 x+4 y-\) \(4=0\) We have \(3 x+4(m x+5-m)-4=0\) \((3+4 m) x+16-4 m=0\) \(\text { So, } x=\frac{4 m-16}{4 m+3} \text { and } y=\frac{m(4 m-16)}{4 m+3}+5-m\) \(\frac{4 m^{2}-16 m-4 m^{2}+17 m+15}{4 m+3}=\frac{m+15}{4 m+3}\) \(\text { So, } Q=\left(\frac{4 m-16}{4 m+3}, \frac{m+15}{4 m+3}\right) \tag{i}\) Since \(\mathrm{m}(15)\) is the mid point of \(\mathrm{PQ}\) we have \(1=\frac{1}{2}\left(\frac{9-m}{5-m}+\frac{4 m-16}{4 m+3}\right) \tag{ii}\) \(\text { and } 5=\frac{1}{2}\left(\frac{25-m}{5-m}+\frac{m+15}{4 m+3}\right) \tag{iii}\) From equation we get \(2(5-m)(4 m+3)=(9-m)(4 m+3)+(5-m)(4 m-16)\) \(2\left(-4 m^{2}+17 m+15\right)=\left(-4 m^{2}+33 m+27\right)+\left(-4 m^{2}+\right.\) \(36 \mathrm{~m}-80)\) \(-8 m^{2}+34 m+30=-8 m^{3}+69 m-53\) \(35 \mathrm{~m}=30+53\) \(\mathrm{m}=\frac{83}{35}\) Putting the value of \(m=83 / 35\) in equation - (i) \(\mathrm{Y}-5=8335(\mathrm{x}-1)\) \(83 \mathrm{x}-35 \mathrm{y}+82=0\) The value of \(m\) obtained from equation (iii) is also equal to \(\mathrm{m}=\frac{83}{35}\)
AP EAMCET-2022-05.07.2022
Co-Ordinate system
88381
If the perpendicular bisector of the line segment joining the points \(P(1,4)\) and \(Q(k, 3)\) has \(y\)-intercept equal to -4 , then the value of \(k\) is
1 \(\sqrt{15}\)
2 -4
3 \(\sqrt{14}\)
4 -2
Explanation:
(B) : The slope of Perpendicular bisector to line PQ \(=-\frac{1}{\text { The slope of the original line PQ }}\) \(=-\frac{1}{\frac{3-4}{\mathrm{k}-1}} \Rightarrow(\mathrm{k}-1)\) The midpoint \(=\left(\frac{\mathrm{k}+1}{2}, \frac{7}{2}\right)\) The equation to the bisector 1 is \(\left(\mathrm{y}-\frac{7}{2}\right)=(\mathrm{k}-1)\left(\mathrm{x}-\frac{\mathrm{k}+1}{2}\right)\) As, \(x=0, y=-4\) satisfies it, we have \(\left(-4-\frac{7}{2}\right)=(\mathrm{k}-1)\left(0-\frac{\mathrm{k}+1}{2}\right)=-\frac{15}{2}=-\frac{\mathrm{k}^{2}-1}{2}\) \(\Rightarrow \mathrm{k}^{2}-1=15\) \(\mathrm{k}^{2}=16\) \(\mathrm{k}= \pm 4\) \(\left(-\infty,-\frac{4}{3}\right) \cup\left(\frac{4}{3}, \infty\right)\)
