88783
If a is parameter then a equation of a family of lines having the sum of the intercepts on axes equal to 7 is
1 \(4 x+3 y=12 a\)
2 \(3 x+4 y=7 a\)
3 \(7 x+a y=a(7-a)\)
4 ay \(=(7-a)(a-x)\)
Explanation:
(D) : Since equation of family of lines is
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
Sum of intercepts \(=a+b=7\) (given)
\(\Rightarrow \mathrm{b}=7-\mathrm{a}\)
Substitute \(b=7-a\) in equation (i), we get
\(\frac{x}{a}+\frac{y}{7-a}=1\)
\((7-a) x+a y=a(7-a)\)
\(a y=(7-a)(a-x)\)
COMEDK-2014
Straight Line
88779
If sum of the slopes of the lines given by \(x^{2}-4 p x y+8 y^{2}=0\) is three times their product, then \(\mathbf{p}=\)
1 3
2 4
3 \(\frac{1}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(D) : Given equation of line,
\(\mathrm{x}^{2}-4 \mathrm{pxy}+8 \mathrm{y}^{2}=0\)
Here \(\mathrm{a}=1,2 \mathrm{~h}=-4 \mathrm{p}, \mathrm{b}=8\)
Let \(m_{1}\) and \(m_{2}\) be the slope of the lines given by eq. (i)
\(\mathrm{m}_{1}+\mathrm{m}_{2}=-\frac{2 \mathrm{~h}}{\mathrm{~b}}=\frac{4 \mathrm{p}}{8}=\frac{\mathrm{p}}{2}\) and \(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{8}\)
\(\mathrm{m}_{1}+\mathrm{m}_{2}=3 \mathrm{~m}_{1} \mathrm{~m}_{2} \Rightarrow \frac{\mathrm{p}}{2}=3\left(\frac{1}{8}\right) \Rightarrow \mathrm{p}=\frac{3}{4}\)
MHT CET-2019
Straight Line
88780
The joint equation of bisectors of angles between lines \(x=5\) and \(y=3\) is
1 \((x-5)(y-3)=0\)
2 \(x^{2}-y^{2}-10 x+6 y+16=0\)
3 \(x y=0\)
4 \(x y-5 x-3 y+15=0\)
Explanation:
(B) :
We can say both line passes through point \((5,3)\) and makes angle \(45^{\circ}\) and \(135^{\circ}\) with \(\mathrm{x}\) axis
\(\therefore\) Equation of \(1^{\text {st }}\) line is,
\(y-3=\tan 45^{\circ}(x-5)\) i.e.
\(\mathrm{y}-3=\mathrm{x}-5\)
\(\mathrm{y}-\mathrm{x}+2=0\)
Equation of \(2^{\text {nd }}\) line is
\(\mathrm{y}-3=\tan 135^{\circ}(\mathrm{x}-5)\) i.e.
\(\mathrm{y}+\mathrm{x}-8=0\)
Required joint equation of line is
\((y-x+2)(y+x-8)=0\)
\(y^{2}-x y+2 y+x y-x^{2}+2 x-8 y+8 x-16=0\)
\(\therefore-x^{2}+y^{2}+10 x-6 y-16=0 \Rightarrow x^{2}-y^{2}-10 x+6 y+16=0\)
MHT CET-2016
Straight Line
88782
If \(4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0\), the family of lines \(a x+b y+c=0\) is concurrent at one or the other of the two points-
1 \(\left(-1,-\frac{1}{2}\right),(-2,-1)\)
2 \((-1,-1),\left(-2,-\frac{1}{2}\right)\)
3 \((-1,2),\left(\frac{1}{2},-1\right)\)
4 \((1,2),\left(\frac{1}{2},-1\right)\)
Explanation:
(A) : Given, the equation of line is
\(4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0\)
\(=(2 \mathrm{a}+\mathrm{b})^{2}-3(2 \mathrm{a}+\mathrm{b}) \mathrm{c}+2 \mathrm{c}^{2}=0\)
\(\Rightarrow(2 \mathrm{a}+\mathrm{b}-2 \mathrm{c})(2 \mathrm{a}+\mathrm{b}-\mathrm{c})=0 \Rightarrow \mathrm{c}=2 \mathrm{a}+\mathrm{b}\)
or \(2 \mathrm{c}=\mathrm{a}+\frac{1}{2} \mathrm{~b}\)
The equation of the family of lines is
\(\mathrm{a}(\mathrm{x}+2)+\mathrm{b}(\mathrm{y}+1)=0\) or \(\mathrm{a}(\mathrm{x}+1)+\mathrm{b}\left(\mathrm{y}+\frac{1}{2}\right)=0\)
The point of concurrent \((-2,-1)\) or \(\left(-1,-\frac{1}{2}\right)\).
BITSAT-2013
Straight Line
88784
If the slope of the lines given by \(\mathrm{ax}^{2}+2 \mathrm{hxy}+\) \(b y^{2}=0\) are in the ratio \(3: 1\), then \(h^{2}\) is equal to
1 \(\frac{\mathrm{ab}}{3}\)
2 \(\frac{4 a b}{3}\)
3 \(\frac{4 \mathrm{a}}{3 \mathrm{~b}}\)
4 None of these
Explanation:
(B) : Given, equation of line,
Sum of Slope \(\left(m_{1}+m_{2}\right)=-\frac{2 h}{b}\)
\(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}\)
Given ratios \(=3: 1\)
\(\mathrm{m}_{1}+3 \mathrm{~m}_{1}=-\frac{2 \mathrm{~h}}{\mathrm{~b}} \tag{i}\)
and \(\quad 3 \mathrm{~m}_{1}^{2}=\frac{\mathrm{a}}{\mathrm{b}}\)
From eq \({ }^{n}\). (i)
\(4 \mathrm{~m}_{1}=-\frac{2 \mathrm{~h}}{\mathrm{~b}}\)
\(\mathrm{~m}_{1}=-\frac{\mathrm{h}}{2 \mathrm{~b}}\)
Substituting value of \(\mathrm{m}_{1}\) in equation (ii) -
\(3 m_{1}^{2}=\frac{a}{b}\)
\(3 \times\left(-\frac{h}{2 b}\right)^{2}=\frac{a}{b}\)
\(3 h^{2}=\frac{4 b^{2} a}{b} \Rightarrow 3 h^{2}=4 a b \Rightarrow h^{2}=\frac{4 a b}{3}\)
88783
If a is parameter then a equation of a family of lines having the sum of the intercepts on axes equal to 7 is
1 \(4 x+3 y=12 a\)
2 \(3 x+4 y=7 a\)
3 \(7 x+a y=a(7-a)\)
4 ay \(=(7-a)(a-x)\)
Explanation:
(D) : Since equation of family of lines is
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
Sum of intercepts \(=a+b=7\) (given)
\(\Rightarrow \mathrm{b}=7-\mathrm{a}\)
Substitute \(b=7-a\) in equation (i), we get
\(\frac{x}{a}+\frac{y}{7-a}=1\)
\((7-a) x+a y=a(7-a)\)
\(a y=(7-a)(a-x)\)
COMEDK-2014
Straight Line
88779
If sum of the slopes of the lines given by \(x^{2}-4 p x y+8 y^{2}=0\) is three times their product, then \(\mathbf{p}=\)
1 3
2 4
3 \(\frac{1}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(D) : Given equation of line,
\(\mathrm{x}^{2}-4 \mathrm{pxy}+8 \mathrm{y}^{2}=0\)
Here \(\mathrm{a}=1,2 \mathrm{~h}=-4 \mathrm{p}, \mathrm{b}=8\)
Let \(m_{1}\) and \(m_{2}\) be the slope of the lines given by eq. (i)
\(\mathrm{m}_{1}+\mathrm{m}_{2}=-\frac{2 \mathrm{~h}}{\mathrm{~b}}=\frac{4 \mathrm{p}}{8}=\frac{\mathrm{p}}{2}\) and \(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{8}\)
\(\mathrm{m}_{1}+\mathrm{m}_{2}=3 \mathrm{~m}_{1} \mathrm{~m}_{2} \Rightarrow \frac{\mathrm{p}}{2}=3\left(\frac{1}{8}\right) \Rightarrow \mathrm{p}=\frac{3}{4}\)
MHT CET-2019
Straight Line
88780
The joint equation of bisectors of angles between lines \(x=5\) and \(y=3\) is
1 \((x-5)(y-3)=0\)
2 \(x^{2}-y^{2}-10 x+6 y+16=0\)
3 \(x y=0\)
4 \(x y-5 x-3 y+15=0\)
Explanation:
(B) :
We can say both line passes through point \((5,3)\) and makes angle \(45^{\circ}\) and \(135^{\circ}\) with \(\mathrm{x}\) axis
\(\therefore\) Equation of \(1^{\text {st }}\) line is,
\(y-3=\tan 45^{\circ}(x-5)\) i.e.
\(\mathrm{y}-3=\mathrm{x}-5\)
\(\mathrm{y}-\mathrm{x}+2=0\)
Equation of \(2^{\text {nd }}\) line is
\(\mathrm{y}-3=\tan 135^{\circ}(\mathrm{x}-5)\) i.e.
\(\mathrm{y}+\mathrm{x}-8=0\)
Required joint equation of line is
\((y-x+2)(y+x-8)=0\)
\(y^{2}-x y+2 y+x y-x^{2}+2 x-8 y+8 x-16=0\)
\(\therefore-x^{2}+y^{2}+10 x-6 y-16=0 \Rightarrow x^{2}-y^{2}-10 x+6 y+16=0\)
MHT CET-2016
Straight Line
88782
If \(4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0\), the family of lines \(a x+b y+c=0\) is concurrent at one or the other of the two points-
1 \(\left(-1,-\frac{1}{2}\right),(-2,-1)\)
2 \((-1,-1),\left(-2,-\frac{1}{2}\right)\)
3 \((-1,2),\left(\frac{1}{2},-1\right)\)
4 \((1,2),\left(\frac{1}{2},-1\right)\)
Explanation:
(A) : Given, the equation of line is
\(4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0\)
\(=(2 \mathrm{a}+\mathrm{b})^{2}-3(2 \mathrm{a}+\mathrm{b}) \mathrm{c}+2 \mathrm{c}^{2}=0\)
\(\Rightarrow(2 \mathrm{a}+\mathrm{b}-2 \mathrm{c})(2 \mathrm{a}+\mathrm{b}-\mathrm{c})=0 \Rightarrow \mathrm{c}=2 \mathrm{a}+\mathrm{b}\)
or \(2 \mathrm{c}=\mathrm{a}+\frac{1}{2} \mathrm{~b}\)
The equation of the family of lines is
\(\mathrm{a}(\mathrm{x}+2)+\mathrm{b}(\mathrm{y}+1)=0\) or \(\mathrm{a}(\mathrm{x}+1)+\mathrm{b}\left(\mathrm{y}+\frac{1}{2}\right)=0\)
The point of concurrent \((-2,-1)\) or \(\left(-1,-\frac{1}{2}\right)\).
BITSAT-2013
Straight Line
88784
If the slope of the lines given by \(\mathrm{ax}^{2}+2 \mathrm{hxy}+\) \(b y^{2}=0\) are in the ratio \(3: 1\), then \(h^{2}\) is equal to
1 \(\frac{\mathrm{ab}}{3}\)
2 \(\frac{4 a b}{3}\)
3 \(\frac{4 \mathrm{a}}{3 \mathrm{~b}}\)
4 None of these
Explanation:
(B) : Given, equation of line,
Sum of Slope \(\left(m_{1}+m_{2}\right)=-\frac{2 h}{b}\)
\(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}\)
Given ratios \(=3: 1\)
\(\mathrm{m}_{1}+3 \mathrm{~m}_{1}=-\frac{2 \mathrm{~h}}{\mathrm{~b}} \tag{i}\)
and \(\quad 3 \mathrm{~m}_{1}^{2}=\frac{\mathrm{a}}{\mathrm{b}}\)
From eq \({ }^{n}\). (i)
\(4 \mathrm{~m}_{1}=-\frac{2 \mathrm{~h}}{\mathrm{~b}}\)
\(\mathrm{~m}_{1}=-\frac{\mathrm{h}}{2 \mathrm{~b}}\)
Substituting value of \(\mathrm{m}_{1}\) in equation (ii) -
\(3 m_{1}^{2}=\frac{a}{b}\)
\(3 \times\left(-\frac{h}{2 b}\right)^{2}=\frac{a}{b}\)
\(3 h^{2}=\frac{4 b^{2} a}{b} \Rightarrow 3 h^{2}=4 a b \Rightarrow h^{2}=\frac{4 a b}{3}\)
88783
If a is parameter then a equation of a family of lines having the sum of the intercepts on axes equal to 7 is
1 \(4 x+3 y=12 a\)
2 \(3 x+4 y=7 a\)
3 \(7 x+a y=a(7-a)\)
4 ay \(=(7-a)(a-x)\)
Explanation:
(D) : Since equation of family of lines is
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
Sum of intercepts \(=a+b=7\) (given)
\(\Rightarrow \mathrm{b}=7-\mathrm{a}\)
Substitute \(b=7-a\) in equation (i), we get
\(\frac{x}{a}+\frac{y}{7-a}=1\)
\((7-a) x+a y=a(7-a)\)
\(a y=(7-a)(a-x)\)
COMEDK-2014
Straight Line
88779
If sum of the slopes of the lines given by \(x^{2}-4 p x y+8 y^{2}=0\) is three times their product, then \(\mathbf{p}=\)
1 3
2 4
3 \(\frac{1}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(D) : Given equation of line,
\(\mathrm{x}^{2}-4 \mathrm{pxy}+8 \mathrm{y}^{2}=0\)
Here \(\mathrm{a}=1,2 \mathrm{~h}=-4 \mathrm{p}, \mathrm{b}=8\)
Let \(m_{1}\) and \(m_{2}\) be the slope of the lines given by eq. (i)
\(\mathrm{m}_{1}+\mathrm{m}_{2}=-\frac{2 \mathrm{~h}}{\mathrm{~b}}=\frac{4 \mathrm{p}}{8}=\frac{\mathrm{p}}{2}\) and \(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{8}\)
\(\mathrm{m}_{1}+\mathrm{m}_{2}=3 \mathrm{~m}_{1} \mathrm{~m}_{2} \Rightarrow \frac{\mathrm{p}}{2}=3\left(\frac{1}{8}\right) \Rightarrow \mathrm{p}=\frac{3}{4}\)
MHT CET-2019
Straight Line
88780
The joint equation of bisectors of angles between lines \(x=5\) and \(y=3\) is
1 \((x-5)(y-3)=0\)
2 \(x^{2}-y^{2}-10 x+6 y+16=0\)
3 \(x y=0\)
4 \(x y-5 x-3 y+15=0\)
Explanation:
(B) :
We can say both line passes through point \((5,3)\) and makes angle \(45^{\circ}\) and \(135^{\circ}\) with \(\mathrm{x}\) axis
\(\therefore\) Equation of \(1^{\text {st }}\) line is,
\(y-3=\tan 45^{\circ}(x-5)\) i.e.
\(\mathrm{y}-3=\mathrm{x}-5\)
\(\mathrm{y}-\mathrm{x}+2=0\)
Equation of \(2^{\text {nd }}\) line is
\(\mathrm{y}-3=\tan 135^{\circ}(\mathrm{x}-5)\) i.e.
\(\mathrm{y}+\mathrm{x}-8=0\)
Required joint equation of line is
\((y-x+2)(y+x-8)=0\)
\(y^{2}-x y+2 y+x y-x^{2}+2 x-8 y+8 x-16=0\)
\(\therefore-x^{2}+y^{2}+10 x-6 y-16=0 \Rightarrow x^{2}-y^{2}-10 x+6 y+16=0\)
MHT CET-2016
Straight Line
88782
If \(4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0\), the family of lines \(a x+b y+c=0\) is concurrent at one or the other of the two points-
1 \(\left(-1,-\frac{1}{2}\right),(-2,-1)\)
2 \((-1,-1),\left(-2,-\frac{1}{2}\right)\)
3 \((-1,2),\left(\frac{1}{2},-1\right)\)
4 \((1,2),\left(\frac{1}{2},-1\right)\)
Explanation:
(A) : Given, the equation of line is
\(4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0\)
\(=(2 \mathrm{a}+\mathrm{b})^{2}-3(2 \mathrm{a}+\mathrm{b}) \mathrm{c}+2 \mathrm{c}^{2}=0\)
\(\Rightarrow(2 \mathrm{a}+\mathrm{b}-2 \mathrm{c})(2 \mathrm{a}+\mathrm{b}-\mathrm{c})=0 \Rightarrow \mathrm{c}=2 \mathrm{a}+\mathrm{b}\)
or \(2 \mathrm{c}=\mathrm{a}+\frac{1}{2} \mathrm{~b}\)
The equation of the family of lines is
\(\mathrm{a}(\mathrm{x}+2)+\mathrm{b}(\mathrm{y}+1)=0\) or \(\mathrm{a}(\mathrm{x}+1)+\mathrm{b}\left(\mathrm{y}+\frac{1}{2}\right)=0\)
The point of concurrent \((-2,-1)\) or \(\left(-1,-\frac{1}{2}\right)\).
BITSAT-2013
Straight Line
88784
If the slope of the lines given by \(\mathrm{ax}^{2}+2 \mathrm{hxy}+\) \(b y^{2}=0\) are in the ratio \(3: 1\), then \(h^{2}\) is equal to
1 \(\frac{\mathrm{ab}}{3}\)
2 \(\frac{4 a b}{3}\)
3 \(\frac{4 \mathrm{a}}{3 \mathrm{~b}}\)
4 None of these
Explanation:
(B) : Given, equation of line,
Sum of Slope \(\left(m_{1}+m_{2}\right)=-\frac{2 h}{b}\)
\(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}\)
Given ratios \(=3: 1\)
\(\mathrm{m}_{1}+3 \mathrm{~m}_{1}=-\frac{2 \mathrm{~h}}{\mathrm{~b}} \tag{i}\)
and \(\quad 3 \mathrm{~m}_{1}^{2}=\frac{\mathrm{a}}{\mathrm{b}}\)
From eq \({ }^{n}\). (i)
\(4 \mathrm{~m}_{1}=-\frac{2 \mathrm{~h}}{\mathrm{~b}}\)
\(\mathrm{~m}_{1}=-\frac{\mathrm{h}}{2 \mathrm{~b}}\)
Substituting value of \(\mathrm{m}_{1}\) in equation (ii) -
\(3 m_{1}^{2}=\frac{a}{b}\)
\(3 \times\left(-\frac{h}{2 b}\right)^{2}=\frac{a}{b}\)
\(3 h^{2}=\frac{4 b^{2} a}{b} \Rightarrow 3 h^{2}=4 a b \Rightarrow h^{2}=\frac{4 a b}{3}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Straight Line
88783
If a is parameter then a equation of a family of lines having the sum of the intercepts on axes equal to 7 is
1 \(4 x+3 y=12 a\)
2 \(3 x+4 y=7 a\)
3 \(7 x+a y=a(7-a)\)
4 ay \(=(7-a)(a-x)\)
Explanation:
(D) : Since equation of family of lines is
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
Sum of intercepts \(=a+b=7\) (given)
\(\Rightarrow \mathrm{b}=7-\mathrm{a}\)
Substitute \(b=7-a\) in equation (i), we get
\(\frac{x}{a}+\frac{y}{7-a}=1\)
\((7-a) x+a y=a(7-a)\)
\(a y=(7-a)(a-x)\)
COMEDK-2014
Straight Line
88779
If sum of the slopes of the lines given by \(x^{2}-4 p x y+8 y^{2}=0\) is three times their product, then \(\mathbf{p}=\)
1 3
2 4
3 \(\frac{1}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(D) : Given equation of line,
\(\mathrm{x}^{2}-4 \mathrm{pxy}+8 \mathrm{y}^{2}=0\)
Here \(\mathrm{a}=1,2 \mathrm{~h}=-4 \mathrm{p}, \mathrm{b}=8\)
Let \(m_{1}\) and \(m_{2}\) be the slope of the lines given by eq. (i)
\(\mathrm{m}_{1}+\mathrm{m}_{2}=-\frac{2 \mathrm{~h}}{\mathrm{~b}}=\frac{4 \mathrm{p}}{8}=\frac{\mathrm{p}}{2}\) and \(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{8}\)
\(\mathrm{m}_{1}+\mathrm{m}_{2}=3 \mathrm{~m}_{1} \mathrm{~m}_{2} \Rightarrow \frac{\mathrm{p}}{2}=3\left(\frac{1}{8}\right) \Rightarrow \mathrm{p}=\frac{3}{4}\)
MHT CET-2019
Straight Line
88780
The joint equation of bisectors of angles between lines \(x=5\) and \(y=3\) is
1 \((x-5)(y-3)=0\)
2 \(x^{2}-y^{2}-10 x+6 y+16=0\)
3 \(x y=0\)
4 \(x y-5 x-3 y+15=0\)
Explanation:
(B) :
We can say both line passes through point \((5,3)\) and makes angle \(45^{\circ}\) and \(135^{\circ}\) with \(\mathrm{x}\) axis
\(\therefore\) Equation of \(1^{\text {st }}\) line is,
\(y-3=\tan 45^{\circ}(x-5)\) i.e.
\(\mathrm{y}-3=\mathrm{x}-5\)
\(\mathrm{y}-\mathrm{x}+2=0\)
Equation of \(2^{\text {nd }}\) line is
\(\mathrm{y}-3=\tan 135^{\circ}(\mathrm{x}-5)\) i.e.
\(\mathrm{y}+\mathrm{x}-8=0\)
Required joint equation of line is
\((y-x+2)(y+x-8)=0\)
\(y^{2}-x y+2 y+x y-x^{2}+2 x-8 y+8 x-16=0\)
\(\therefore-x^{2}+y^{2}+10 x-6 y-16=0 \Rightarrow x^{2}-y^{2}-10 x+6 y+16=0\)
MHT CET-2016
Straight Line
88782
If \(4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0\), the family of lines \(a x+b y+c=0\) is concurrent at one or the other of the two points-
1 \(\left(-1,-\frac{1}{2}\right),(-2,-1)\)
2 \((-1,-1),\left(-2,-\frac{1}{2}\right)\)
3 \((-1,2),\left(\frac{1}{2},-1\right)\)
4 \((1,2),\left(\frac{1}{2},-1\right)\)
Explanation:
(A) : Given, the equation of line is
\(4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0\)
\(=(2 \mathrm{a}+\mathrm{b})^{2}-3(2 \mathrm{a}+\mathrm{b}) \mathrm{c}+2 \mathrm{c}^{2}=0\)
\(\Rightarrow(2 \mathrm{a}+\mathrm{b}-2 \mathrm{c})(2 \mathrm{a}+\mathrm{b}-\mathrm{c})=0 \Rightarrow \mathrm{c}=2 \mathrm{a}+\mathrm{b}\)
or \(2 \mathrm{c}=\mathrm{a}+\frac{1}{2} \mathrm{~b}\)
The equation of the family of lines is
\(\mathrm{a}(\mathrm{x}+2)+\mathrm{b}(\mathrm{y}+1)=0\) or \(\mathrm{a}(\mathrm{x}+1)+\mathrm{b}\left(\mathrm{y}+\frac{1}{2}\right)=0\)
The point of concurrent \((-2,-1)\) or \(\left(-1,-\frac{1}{2}\right)\).
BITSAT-2013
Straight Line
88784
If the slope of the lines given by \(\mathrm{ax}^{2}+2 \mathrm{hxy}+\) \(b y^{2}=0\) are in the ratio \(3: 1\), then \(h^{2}\) is equal to
1 \(\frac{\mathrm{ab}}{3}\)
2 \(\frac{4 a b}{3}\)
3 \(\frac{4 \mathrm{a}}{3 \mathrm{~b}}\)
4 None of these
Explanation:
(B) : Given, equation of line,
Sum of Slope \(\left(m_{1}+m_{2}\right)=-\frac{2 h}{b}\)
\(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}\)
Given ratios \(=3: 1\)
\(\mathrm{m}_{1}+3 \mathrm{~m}_{1}=-\frac{2 \mathrm{~h}}{\mathrm{~b}} \tag{i}\)
and \(\quad 3 \mathrm{~m}_{1}^{2}=\frac{\mathrm{a}}{\mathrm{b}}\)
From eq \({ }^{n}\). (i)
\(4 \mathrm{~m}_{1}=-\frac{2 \mathrm{~h}}{\mathrm{~b}}\)
\(\mathrm{~m}_{1}=-\frac{\mathrm{h}}{2 \mathrm{~b}}\)
Substituting value of \(\mathrm{m}_{1}\) in equation (ii) -
\(3 m_{1}^{2}=\frac{a}{b}\)
\(3 \times\left(-\frac{h}{2 b}\right)^{2}=\frac{a}{b}\)
\(3 h^{2}=\frac{4 b^{2} a}{b} \Rightarrow 3 h^{2}=4 a b \Rightarrow h^{2}=\frac{4 a b}{3}\)
88783
If a is parameter then a equation of a family of lines having the sum of the intercepts on axes equal to 7 is
1 \(4 x+3 y=12 a\)
2 \(3 x+4 y=7 a\)
3 \(7 x+a y=a(7-a)\)
4 ay \(=(7-a)(a-x)\)
Explanation:
(D) : Since equation of family of lines is
\(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\)
Sum of intercepts \(=a+b=7\) (given)
\(\Rightarrow \mathrm{b}=7-\mathrm{a}\)
Substitute \(b=7-a\) in equation (i), we get
\(\frac{x}{a}+\frac{y}{7-a}=1\)
\((7-a) x+a y=a(7-a)\)
\(a y=(7-a)(a-x)\)
COMEDK-2014
Straight Line
88779
If sum of the slopes of the lines given by \(x^{2}-4 p x y+8 y^{2}=0\) is three times their product, then \(\mathbf{p}=\)
1 3
2 4
3 \(\frac{1}{4}\)
4 \(\frac{3}{4}\)
Explanation:
(D) : Given equation of line,
\(\mathrm{x}^{2}-4 \mathrm{pxy}+8 \mathrm{y}^{2}=0\)
Here \(\mathrm{a}=1,2 \mathrm{~h}=-4 \mathrm{p}, \mathrm{b}=8\)
Let \(m_{1}\) and \(m_{2}\) be the slope of the lines given by eq. (i)
\(\mathrm{m}_{1}+\mathrm{m}_{2}=-\frac{2 \mathrm{~h}}{\mathrm{~b}}=\frac{4 \mathrm{p}}{8}=\frac{\mathrm{p}}{2}\) and \(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{8}\)
\(\mathrm{m}_{1}+\mathrm{m}_{2}=3 \mathrm{~m}_{1} \mathrm{~m}_{2} \Rightarrow \frac{\mathrm{p}}{2}=3\left(\frac{1}{8}\right) \Rightarrow \mathrm{p}=\frac{3}{4}\)
MHT CET-2019
Straight Line
88780
The joint equation of bisectors of angles between lines \(x=5\) and \(y=3\) is
1 \((x-5)(y-3)=0\)
2 \(x^{2}-y^{2}-10 x+6 y+16=0\)
3 \(x y=0\)
4 \(x y-5 x-3 y+15=0\)
Explanation:
(B) :
We can say both line passes through point \((5,3)\) and makes angle \(45^{\circ}\) and \(135^{\circ}\) with \(\mathrm{x}\) axis
\(\therefore\) Equation of \(1^{\text {st }}\) line is,
\(y-3=\tan 45^{\circ}(x-5)\) i.e.
\(\mathrm{y}-3=\mathrm{x}-5\)
\(\mathrm{y}-\mathrm{x}+2=0\)
Equation of \(2^{\text {nd }}\) line is
\(\mathrm{y}-3=\tan 135^{\circ}(\mathrm{x}-5)\) i.e.
\(\mathrm{y}+\mathrm{x}-8=0\)
Required joint equation of line is
\((y-x+2)(y+x-8)=0\)
\(y^{2}-x y+2 y+x y-x^{2}+2 x-8 y+8 x-16=0\)
\(\therefore-x^{2}+y^{2}+10 x-6 y-16=0 \Rightarrow x^{2}-y^{2}-10 x+6 y+16=0\)
MHT CET-2016
Straight Line
88782
If \(4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0\), the family of lines \(a x+b y+c=0\) is concurrent at one or the other of the two points-
1 \(\left(-1,-\frac{1}{2}\right),(-2,-1)\)
2 \((-1,-1),\left(-2,-\frac{1}{2}\right)\)
3 \((-1,2),\left(\frac{1}{2},-1\right)\)
4 \((1,2),\left(\frac{1}{2},-1\right)\)
Explanation:
(A) : Given, the equation of line is
\(4 a^{2}+b^{2}+2 c^{2}+4 a b-6 a c-3 b c=0\)
\(=(2 \mathrm{a}+\mathrm{b})^{2}-3(2 \mathrm{a}+\mathrm{b}) \mathrm{c}+2 \mathrm{c}^{2}=0\)
\(\Rightarrow(2 \mathrm{a}+\mathrm{b}-2 \mathrm{c})(2 \mathrm{a}+\mathrm{b}-\mathrm{c})=0 \Rightarrow \mathrm{c}=2 \mathrm{a}+\mathrm{b}\)
or \(2 \mathrm{c}=\mathrm{a}+\frac{1}{2} \mathrm{~b}\)
The equation of the family of lines is
\(\mathrm{a}(\mathrm{x}+2)+\mathrm{b}(\mathrm{y}+1)=0\) or \(\mathrm{a}(\mathrm{x}+1)+\mathrm{b}\left(\mathrm{y}+\frac{1}{2}\right)=0\)
The point of concurrent \((-2,-1)\) or \(\left(-1,-\frac{1}{2}\right)\).
BITSAT-2013
Straight Line
88784
If the slope of the lines given by \(\mathrm{ax}^{2}+2 \mathrm{hxy}+\) \(b y^{2}=0\) are in the ratio \(3: 1\), then \(h^{2}\) is equal to
1 \(\frac{\mathrm{ab}}{3}\)
2 \(\frac{4 a b}{3}\)
3 \(\frac{4 \mathrm{a}}{3 \mathrm{~b}}\)
4 None of these
Explanation:
(B) : Given, equation of line,
Sum of Slope \(\left(m_{1}+m_{2}\right)=-\frac{2 h}{b}\)
\(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}\)
Given ratios \(=3: 1\)
\(\mathrm{m}_{1}+3 \mathrm{~m}_{1}=-\frac{2 \mathrm{~h}}{\mathrm{~b}} \tag{i}\)
and \(\quad 3 \mathrm{~m}_{1}^{2}=\frac{\mathrm{a}}{\mathrm{b}}\)
From eq \({ }^{n}\). (i)
\(4 \mathrm{~m}_{1}=-\frac{2 \mathrm{~h}}{\mathrm{~b}}\)
\(\mathrm{~m}_{1}=-\frac{\mathrm{h}}{2 \mathrm{~b}}\)
Substituting value of \(\mathrm{m}_{1}\) in equation (ii) -
\(3 m_{1}^{2}=\frac{a}{b}\)
\(3 \times\left(-\frac{h}{2 b}\right)^{2}=\frac{a}{b}\)
\(3 h^{2}=\frac{4 b^{2} a}{b} \Rightarrow 3 h^{2}=4 a b \Rightarrow h^{2}=\frac{4 a b}{3}\)