88206
If the straight lines \(2 x+3 y-3=0\) and \(x+k y+\) \(\mathbf{7}=\mathbf{0}\) are perpendicular, then the value of \(k\) is
1 \(2 / 3\)
2 \(3 / 2\)
3 \(-2 / 3\)
4 \(-3 / 2\)
Explanation:
(C) : Given, Equation of straight lines are \(\begin{align*} & 2 x+3 y-3=0 \tag{i}\\ & x+k y+7=0 \tag{ii} \end{align*}\) Equation (i) and (ii) can be written as shown below \(y=-\frac{2}{3} x+1\) \(\therefore\) Slope \(\left(\mathrm{m}_{1}\right)=-\frac{2}{3}\) And, \(\quad \mathrm{y}=-\frac{1}{\mathrm{k}} \mathrm{x}-\frac{7}{\mathrm{k}}\) Slope \(\left(\mathrm{m}_{2}\right)=-\frac{1}{\mathrm{k}}\) We know that both the straight line are perpendicular to each other. \(\therefore \quad \mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(-\frac{2}{3} \times\left(-\frac{1}{\mathrm{k}}\right)=-1, \quad \mathrm{k}=-\frac{2}{3}\)
Karnataka CET-2016
Co-Ordinate system
88207
A straight line passes through the points \((5,0)\) and \((0,3)\). The length of perpendicular from the point \((4,4)\) on the line is
1 \(\frac{15}{\sqrt{34}}\)
2 \(\frac{\sqrt{17}}{2}\)
3 \(\frac{17}{2}\)
4 \(\sqrt{\frac{17}{2}}\)
Explanation:
(D) : We know, The equation Straight line passes through point \((5,0)\) and \((0,3)\) \(\left(y-y_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)\) \((y-0)=\frac{3-0}{0-5}(x-5)\) \(y=-\frac{3}{5}(x-5)\) \(5 y=-3 x+15\) \(3 x+5 y-15=0\) Therefore, the distance of this line from the point \((4,4)\) is \(d=\frac{|3 \times 4+5 \times 4-15|}{\sqrt{3^{2}+5^{2}}}=\frac{17}{\sqrt{34}}\) \(d=\frac{\sqrt{17} \times \sqrt{17}}{\sqrt{17} \times \sqrt{2}}=\sqrt{\frac{17}{2}}\)
Karnataka CET-2014
Co-Ordinate system
88208
Find the transformed equation of the straight line \(x y-x-y+1=0\), when the origin is shifted to the point \((1,1)\) after translation of axis.
1 \(x y=5\)
2 \(x y=2\)
3 \(x y=0\)
4 \(x y=8\)
Explanation:
(C) : Let the coordinates of a point \(p\) changes from \((x, y)\) to \(\left(x^{\prime}, y^{\prime}\right)\) in new coordinates axis where origin has the coordinates \(\mathrm{h}=1, \mathrm{k}=1\) Then, \(\mathrm{x}=\mathrm{x}^{\prime}+1, \mathrm{y}=\mathrm{y}^{\prime}+1\). Substituting these values in the given equation of straight line \(\left(x^{\prime}+1\right)\left(y^{\prime}+1\right)-\left(x^{\prime}+1\right)-\left(y^{\prime}+1\right)+1=0\) \(x^{\prime} y^{\prime}+x^{\prime}+y^{\prime}+1-x^{\prime}-1-y^{\prime}-1+1=0\) \(x^{\prime} y^{\prime}=0\) Therefore, the equation of straight line in the new system is \(\mathrm{xy}=0\).
COMEDK-2017
Co-Ordinate system
88209
If the axis are shifted to the point \((1,-2)\) without solution, then the equation \(2 x^{2}+y^{2}-4 x+4 y=0\) becomes
1 \(2 \mathrm{X}^{2}+3 \mathrm{Y}^{2}=6\)
2 \(2 \mathrm{X}^{2}+\mathrm{Y}^{2}=6\)
3 \(\mathrm{X}^{2}+2 \mathrm{Y}^{2}=6\)
4 None of these
Explanation:
(B) :Given the point \((1,-2)\) The equation is \(2 x^{2}+y^{2}-4 x+4 y=0\) Now, substituting \(\mathrm{x}=\mathrm{X}+1\) and \(\mathrm{y}=\mathrm{Y}-2\) in given equation, we get \(2(X+1)^{2}+(Y-2)^{2}-4(X+1)+4(Y-2)=0\) \(2 X^{2}+Y^{2}=6\)
88206
If the straight lines \(2 x+3 y-3=0\) and \(x+k y+\) \(\mathbf{7}=\mathbf{0}\) are perpendicular, then the value of \(k\) is
1 \(2 / 3\)
2 \(3 / 2\)
3 \(-2 / 3\)
4 \(-3 / 2\)
Explanation:
(C) : Given, Equation of straight lines are \(\begin{align*} & 2 x+3 y-3=0 \tag{i}\\ & x+k y+7=0 \tag{ii} \end{align*}\) Equation (i) and (ii) can be written as shown below \(y=-\frac{2}{3} x+1\) \(\therefore\) Slope \(\left(\mathrm{m}_{1}\right)=-\frac{2}{3}\) And, \(\quad \mathrm{y}=-\frac{1}{\mathrm{k}} \mathrm{x}-\frac{7}{\mathrm{k}}\) Slope \(\left(\mathrm{m}_{2}\right)=-\frac{1}{\mathrm{k}}\) We know that both the straight line are perpendicular to each other. \(\therefore \quad \mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(-\frac{2}{3} \times\left(-\frac{1}{\mathrm{k}}\right)=-1, \quad \mathrm{k}=-\frac{2}{3}\)
Karnataka CET-2016
Co-Ordinate system
88207
A straight line passes through the points \((5,0)\) and \((0,3)\). The length of perpendicular from the point \((4,4)\) on the line is
1 \(\frac{15}{\sqrt{34}}\)
2 \(\frac{\sqrt{17}}{2}\)
3 \(\frac{17}{2}\)
4 \(\sqrt{\frac{17}{2}}\)
Explanation:
(D) : We know, The equation Straight line passes through point \((5,0)\) and \((0,3)\) \(\left(y-y_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)\) \((y-0)=\frac{3-0}{0-5}(x-5)\) \(y=-\frac{3}{5}(x-5)\) \(5 y=-3 x+15\) \(3 x+5 y-15=0\) Therefore, the distance of this line from the point \((4,4)\) is \(d=\frac{|3 \times 4+5 \times 4-15|}{\sqrt{3^{2}+5^{2}}}=\frac{17}{\sqrt{34}}\) \(d=\frac{\sqrt{17} \times \sqrt{17}}{\sqrt{17} \times \sqrt{2}}=\sqrt{\frac{17}{2}}\)
Karnataka CET-2014
Co-Ordinate system
88208
Find the transformed equation of the straight line \(x y-x-y+1=0\), when the origin is shifted to the point \((1,1)\) after translation of axis.
1 \(x y=5\)
2 \(x y=2\)
3 \(x y=0\)
4 \(x y=8\)
Explanation:
(C) : Let the coordinates of a point \(p\) changes from \((x, y)\) to \(\left(x^{\prime}, y^{\prime}\right)\) in new coordinates axis where origin has the coordinates \(\mathrm{h}=1, \mathrm{k}=1\) Then, \(\mathrm{x}=\mathrm{x}^{\prime}+1, \mathrm{y}=\mathrm{y}^{\prime}+1\). Substituting these values in the given equation of straight line \(\left(x^{\prime}+1\right)\left(y^{\prime}+1\right)-\left(x^{\prime}+1\right)-\left(y^{\prime}+1\right)+1=0\) \(x^{\prime} y^{\prime}+x^{\prime}+y^{\prime}+1-x^{\prime}-1-y^{\prime}-1+1=0\) \(x^{\prime} y^{\prime}=0\) Therefore, the equation of straight line in the new system is \(\mathrm{xy}=0\).
COMEDK-2017
Co-Ordinate system
88209
If the axis are shifted to the point \((1,-2)\) without solution, then the equation \(2 x^{2}+y^{2}-4 x+4 y=0\) becomes
1 \(2 \mathrm{X}^{2}+3 \mathrm{Y}^{2}=6\)
2 \(2 \mathrm{X}^{2}+\mathrm{Y}^{2}=6\)
3 \(\mathrm{X}^{2}+2 \mathrm{Y}^{2}=6\)
4 None of these
Explanation:
(B) :Given the point \((1,-2)\) The equation is \(2 x^{2}+y^{2}-4 x+4 y=0\) Now, substituting \(\mathrm{x}=\mathrm{X}+1\) and \(\mathrm{y}=\mathrm{Y}-2\) in given equation, we get \(2(X+1)^{2}+(Y-2)^{2}-4(X+1)+4(Y-2)=0\) \(2 X^{2}+Y^{2}=6\)
88206
If the straight lines \(2 x+3 y-3=0\) and \(x+k y+\) \(\mathbf{7}=\mathbf{0}\) are perpendicular, then the value of \(k\) is
1 \(2 / 3\)
2 \(3 / 2\)
3 \(-2 / 3\)
4 \(-3 / 2\)
Explanation:
(C) : Given, Equation of straight lines are \(\begin{align*} & 2 x+3 y-3=0 \tag{i}\\ & x+k y+7=0 \tag{ii} \end{align*}\) Equation (i) and (ii) can be written as shown below \(y=-\frac{2}{3} x+1\) \(\therefore\) Slope \(\left(\mathrm{m}_{1}\right)=-\frac{2}{3}\) And, \(\quad \mathrm{y}=-\frac{1}{\mathrm{k}} \mathrm{x}-\frac{7}{\mathrm{k}}\) Slope \(\left(\mathrm{m}_{2}\right)=-\frac{1}{\mathrm{k}}\) We know that both the straight line are perpendicular to each other. \(\therefore \quad \mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(-\frac{2}{3} \times\left(-\frac{1}{\mathrm{k}}\right)=-1, \quad \mathrm{k}=-\frac{2}{3}\)
Karnataka CET-2016
Co-Ordinate system
88207
A straight line passes through the points \((5,0)\) and \((0,3)\). The length of perpendicular from the point \((4,4)\) on the line is
1 \(\frac{15}{\sqrt{34}}\)
2 \(\frac{\sqrt{17}}{2}\)
3 \(\frac{17}{2}\)
4 \(\sqrt{\frac{17}{2}}\)
Explanation:
(D) : We know, The equation Straight line passes through point \((5,0)\) and \((0,3)\) \(\left(y-y_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)\) \((y-0)=\frac{3-0}{0-5}(x-5)\) \(y=-\frac{3}{5}(x-5)\) \(5 y=-3 x+15\) \(3 x+5 y-15=0\) Therefore, the distance of this line from the point \((4,4)\) is \(d=\frac{|3 \times 4+5 \times 4-15|}{\sqrt{3^{2}+5^{2}}}=\frac{17}{\sqrt{34}}\) \(d=\frac{\sqrt{17} \times \sqrt{17}}{\sqrt{17} \times \sqrt{2}}=\sqrt{\frac{17}{2}}\)
Karnataka CET-2014
Co-Ordinate system
88208
Find the transformed equation of the straight line \(x y-x-y+1=0\), when the origin is shifted to the point \((1,1)\) after translation of axis.
1 \(x y=5\)
2 \(x y=2\)
3 \(x y=0\)
4 \(x y=8\)
Explanation:
(C) : Let the coordinates of a point \(p\) changes from \((x, y)\) to \(\left(x^{\prime}, y^{\prime}\right)\) in new coordinates axis where origin has the coordinates \(\mathrm{h}=1, \mathrm{k}=1\) Then, \(\mathrm{x}=\mathrm{x}^{\prime}+1, \mathrm{y}=\mathrm{y}^{\prime}+1\). Substituting these values in the given equation of straight line \(\left(x^{\prime}+1\right)\left(y^{\prime}+1\right)-\left(x^{\prime}+1\right)-\left(y^{\prime}+1\right)+1=0\) \(x^{\prime} y^{\prime}+x^{\prime}+y^{\prime}+1-x^{\prime}-1-y^{\prime}-1+1=0\) \(x^{\prime} y^{\prime}=0\) Therefore, the equation of straight line in the new system is \(\mathrm{xy}=0\).
COMEDK-2017
Co-Ordinate system
88209
If the axis are shifted to the point \((1,-2)\) without solution, then the equation \(2 x^{2}+y^{2}-4 x+4 y=0\) becomes
1 \(2 \mathrm{X}^{2}+3 \mathrm{Y}^{2}=6\)
2 \(2 \mathrm{X}^{2}+\mathrm{Y}^{2}=6\)
3 \(\mathrm{X}^{2}+2 \mathrm{Y}^{2}=6\)
4 None of these
Explanation:
(B) :Given the point \((1,-2)\) The equation is \(2 x^{2}+y^{2}-4 x+4 y=0\) Now, substituting \(\mathrm{x}=\mathrm{X}+1\) and \(\mathrm{y}=\mathrm{Y}-2\) in given equation, we get \(2(X+1)^{2}+(Y-2)^{2}-4(X+1)+4(Y-2)=0\) \(2 X^{2}+Y^{2}=6\)
88206
If the straight lines \(2 x+3 y-3=0\) and \(x+k y+\) \(\mathbf{7}=\mathbf{0}\) are perpendicular, then the value of \(k\) is
1 \(2 / 3\)
2 \(3 / 2\)
3 \(-2 / 3\)
4 \(-3 / 2\)
Explanation:
(C) : Given, Equation of straight lines are \(\begin{align*} & 2 x+3 y-3=0 \tag{i}\\ & x+k y+7=0 \tag{ii} \end{align*}\) Equation (i) and (ii) can be written as shown below \(y=-\frac{2}{3} x+1\) \(\therefore\) Slope \(\left(\mathrm{m}_{1}\right)=-\frac{2}{3}\) And, \(\quad \mathrm{y}=-\frac{1}{\mathrm{k}} \mathrm{x}-\frac{7}{\mathrm{k}}\) Slope \(\left(\mathrm{m}_{2}\right)=-\frac{1}{\mathrm{k}}\) We know that both the straight line are perpendicular to each other. \(\therefore \quad \mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(-\frac{2}{3} \times\left(-\frac{1}{\mathrm{k}}\right)=-1, \quad \mathrm{k}=-\frac{2}{3}\)
Karnataka CET-2016
Co-Ordinate system
88207
A straight line passes through the points \((5,0)\) and \((0,3)\). The length of perpendicular from the point \((4,4)\) on the line is
1 \(\frac{15}{\sqrt{34}}\)
2 \(\frac{\sqrt{17}}{2}\)
3 \(\frac{17}{2}\)
4 \(\sqrt{\frac{17}{2}}\)
Explanation:
(D) : We know, The equation Straight line passes through point \((5,0)\) and \((0,3)\) \(\left(y-y_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)\) \((y-0)=\frac{3-0}{0-5}(x-5)\) \(y=-\frac{3}{5}(x-5)\) \(5 y=-3 x+15\) \(3 x+5 y-15=0\) Therefore, the distance of this line from the point \((4,4)\) is \(d=\frac{|3 \times 4+5 \times 4-15|}{\sqrt{3^{2}+5^{2}}}=\frac{17}{\sqrt{34}}\) \(d=\frac{\sqrt{17} \times \sqrt{17}}{\sqrt{17} \times \sqrt{2}}=\sqrt{\frac{17}{2}}\)
Karnataka CET-2014
Co-Ordinate system
88208
Find the transformed equation of the straight line \(x y-x-y+1=0\), when the origin is shifted to the point \((1,1)\) after translation of axis.
1 \(x y=5\)
2 \(x y=2\)
3 \(x y=0\)
4 \(x y=8\)
Explanation:
(C) : Let the coordinates of a point \(p\) changes from \((x, y)\) to \(\left(x^{\prime}, y^{\prime}\right)\) in new coordinates axis where origin has the coordinates \(\mathrm{h}=1, \mathrm{k}=1\) Then, \(\mathrm{x}=\mathrm{x}^{\prime}+1, \mathrm{y}=\mathrm{y}^{\prime}+1\). Substituting these values in the given equation of straight line \(\left(x^{\prime}+1\right)\left(y^{\prime}+1\right)-\left(x^{\prime}+1\right)-\left(y^{\prime}+1\right)+1=0\) \(x^{\prime} y^{\prime}+x^{\prime}+y^{\prime}+1-x^{\prime}-1-y^{\prime}-1+1=0\) \(x^{\prime} y^{\prime}=0\) Therefore, the equation of straight line in the new system is \(\mathrm{xy}=0\).
COMEDK-2017
Co-Ordinate system
88209
If the axis are shifted to the point \((1,-2)\) without solution, then the equation \(2 x^{2}+y^{2}-4 x+4 y=0\) becomes
1 \(2 \mathrm{X}^{2}+3 \mathrm{Y}^{2}=6\)
2 \(2 \mathrm{X}^{2}+\mathrm{Y}^{2}=6\)
3 \(\mathrm{X}^{2}+2 \mathrm{Y}^{2}=6\)
4 None of these
Explanation:
(B) :Given the point \((1,-2)\) The equation is \(2 x^{2}+y^{2}-4 x+4 y=0\) Now, substituting \(\mathrm{x}=\mathrm{X}+1\) and \(\mathrm{y}=\mathrm{Y}-2\) in given equation, we get \(2(X+1)^{2}+(Y-2)^{2}-4(X+1)+4(Y-2)=0\) \(2 X^{2}+Y^{2}=6\)
