88642
The line through the points \((1,4),(-5,1)\) intersects the line \(4 x+3 y-5=0\) in the point
1 \(\left(\frac{5}{3}, \frac{-5}{3}\right)\)
2 \((-1,-3)\)
3 \((2,1)\)
4 \(-1,3)\)
Explanation:
(D) :
Equation of line passing through \(\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(1,4)\) and \(\mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(-5,1)\) is
\(\frac{y-4}{4-1}=\frac{x-1}{1+5}\)
\(\frac{y-4}{3}=\frac{x-1}{6}\)
\(\Rightarrow 2(y-4)=x-1\)
\(\Rightarrow x-1-2 y+8=0\)
\(x-2 y+7=0 \tag{i}\)
Also given equation of line is
\(4 x+3 y-5=0 \tag{ii}\)
These two line intersect each other.
So, solving (i) and (ii) we get \(x=-1, y=3\)
MHT CET-2020
Straight Line
88644
The equation of the line parallel to the line \(3 x-\) \(4 y+2=0\) and passing through \((-2,3)\) is
1 \(3 x-4 y+18=0\)
2 \(3 x-4 y-18=0\)
3 \(3 x+4 y+18=0\)
4 \(3 x+4 y-18=0\)
Explanation:
(A) : Given the equation of the line, \(3 x-4 y+2=0\) \(4 y=3 x+2\)
\(y=\left(\frac{3}{4}\right) x+\frac{1}{2}\)
\(\therefore\) The slope of given line is \(\left(\frac{3}{4}\right)\)
slope of given line \(=\) slope of parallel line \(=\frac{3}{4}\)
Equation of line passes through \((-2,3)\) with slope \(\left(\frac{3}{4}\right)\) is
\((y-3)=\frac{3}{4}(x+2)\)
\(4 y-12=3 x+6\)
\(3 x-4 y+18=0\)
Karnataka CET-2018
Straight Line
88645
If lines represented by \(x+3 y-6=0,2 x+y-4\) \(=0\) and \(k x-3 y+1=0\) are concurrent, then the value of \(k\) is
88646
If the lines \((p-q) x^{2}+2(p+q) x y+(q-p) y^{2}=\) 0 are mutually perpendicular, then
1 \(p=q\)
2 \(\mathrm{p}=0\)
3 \(q=0\)
4 p and q may have any value
Explanation:
(D) : Given, \((p-q) x^{2}+2(p+q) x y+(q-p) y^{2}=0\)
\(\because\) these lines are perpendicular to each other.
\(\therefore\) Coefficient of \(\mathrm{x}^{2}+\) Coefficient of \(\mathrm{y}^{2}=0\)
\(\Rightarrow \mathrm{p}-\mathrm{q}+\mathrm{q}-\mathrm{p}=0\)
\(\therefore \mathrm{p}\) and \(\mathrm{q}\) may have any value.
88642
The line through the points \((1,4),(-5,1)\) intersects the line \(4 x+3 y-5=0\) in the point
1 \(\left(\frac{5}{3}, \frac{-5}{3}\right)\)
2 \((-1,-3)\)
3 \((2,1)\)
4 \(-1,3)\)
Explanation:
(D) :
Equation of line passing through \(\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(1,4)\) and \(\mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(-5,1)\) is
\(\frac{y-4}{4-1}=\frac{x-1}{1+5}\)
\(\frac{y-4}{3}=\frac{x-1}{6}\)
\(\Rightarrow 2(y-4)=x-1\)
\(\Rightarrow x-1-2 y+8=0\)
\(x-2 y+7=0 \tag{i}\)
Also given equation of line is
\(4 x+3 y-5=0 \tag{ii}\)
These two line intersect each other.
So, solving (i) and (ii) we get \(x=-1, y=3\)
MHT CET-2020
Straight Line
88644
The equation of the line parallel to the line \(3 x-\) \(4 y+2=0\) and passing through \((-2,3)\) is
1 \(3 x-4 y+18=0\)
2 \(3 x-4 y-18=0\)
3 \(3 x+4 y+18=0\)
4 \(3 x+4 y-18=0\)
Explanation:
(A) : Given the equation of the line, \(3 x-4 y+2=0\) \(4 y=3 x+2\)
\(y=\left(\frac{3}{4}\right) x+\frac{1}{2}\)
\(\therefore\) The slope of given line is \(\left(\frac{3}{4}\right)\)
slope of given line \(=\) slope of parallel line \(=\frac{3}{4}\)
Equation of line passes through \((-2,3)\) with slope \(\left(\frac{3}{4}\right)\) is
\((y-3)=\frac{3}{4}(x+2)\)
\(4 y-12=3 x+6\)
\(3 x-4 y+18=0\)
Karnataka CET-2018
Straight Line
88645
If lines represented by \(x+3 y-6=0,2 x+y-4\) \(=0\) and \(k x-3 y+1=0\) are concurrent, then the value of \(k\) is
88646
If the lines \((p-q) x^{2}+2(p+q) x y+(q-p) y^{2}=\) 0 are mutually perpendicular, then
1 \(p=q\)
2 \(\mathrm{p}=0\)
3 \(q=0\)
4 p and q may have any value
Explanation:
(D) : Given, \((p-q) x^{2}+2(p+q) x y+(q-p) y^{2}=0\)
\(\because\) these lines are perpendicular to each other.
\(\therefore\) Coefficient of \(\mathrm{x}^{2}+\) Coefficient of \(\mathrm{y}^{2}=0\)
\(\Rightarrow \mathrm{p}-\mathrm{q}+\mathrm{q}-\mathrm{p}=0\)
\(\therefore \mathrm{p}\) and \(\mathrm{q}\) may have any value.
88642
The line through the points \((1,4),(-5,1)\) intersects the line \(4 x+3 y-5=0\) in the point
1 \(\left(\frac{5}{3}, \frac{-5}{3}\right)\)
2 \((-1,-3)\)
3 \((2,1)\)
4 \(-1,3)\)
Explanation:
(D) :
Equation of line passing through \(\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(1,4)\) and \(\mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(-5,1)\) is
\(\frac{y-4}{4-1}=\frac{x-1}{1+5}\)
\(\frac{y-4}{3}=\frac{x-1}{6}\)
\(\Rightarrow 2(y-4)=x-1\)
\(\Rightarrow x-1-2 y+8=0\)
\(x-2 y+7=0 \tag{i}\)
Also given equation of line is
\(4 x+3 y-5=0 \tag{ii}\)
These two line intersect each other.
So, solving (i) and (ii) we get \(x=-1, y=3\)
MHT CET-2020
Straight Line
88644
The equation of the line parallel to the line \(3 x-\) \(4 y+2=0\) and passing through \((-2,3)\) is
1 \(3 x-4 y+18=0\)
2 \(3 x-4 y-18=0\)
3 \(3 x+4 y+18=0\)
4 \(3 x+4 y-18=0\)
Explanation:
(A) : Given the equation of the line, \(3 x-4 y+2=0\) \(4 y=3 x+2\)
\(y=\left(\frac{3}{4}\right) x+\frac{1}{2}\)
\(\therefore\) The slope of given line is \(\left(\frac{3}{4}\right)\)
slope of given line \(=\) slope of parallel line \(=\frac{3}{4}\)
Equation of line passes through \((-2,3)\) with slope \(\left(\frac{3}{4}\right)\) is
\((y-3)=\frac{3}{4}(x+2)\)
\(4 y-12=3 x+6\)
\(3 x-4 y+18=0\)
Karnataka CET-2018
Straight Line
88645
If lines represented by \(x+3 y-6=0,2 x+y-4\) \(=0\) and \(k x-3 y+1=0\) are concurrent, then the value of \(k\) is
88646
If the lines \((p-q) x^{2}+2(p+q) x y+(q-p) y^{2}=\) 0 are mutually perpendicular, then
1 \(p=q\)
2 \(\mathrm{p}=0\)
3 \(q=0\)
4 p and q may have any value
Explanation:
(D) : Given, \((p-q) x^{2}+2(p+q) x y+(q-p) y^{2}=0\)
\(\because\) these lines are perpendicular to each other.
\(\therefore\) Coefficient of \(\mathrm{x}^{2}+\) Coefficient of \(\mathrm{y}^{2}=0\)
\(\Rightarrow \mathrm{p}-\mathrm{q}+\mathrm{q}-\mathrm{p}=0\)
\(\therefore \mathrm{p}\) and \(\mathrm{q}\) may have any value.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Straight Line
88642
The line through the points \((1,4),(-5,1)\) intersects the line \(4 x+3 y-5=0\) in the point
1 \(\left(\frac{5}{3}, \frac{-5}{3}\right)\)
2 \((-1,-3)\)
3 \((2,1)\)
4 \(-1,3)\)
Explanation:
(D) :
Equation of line passing through \(\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(1,4)\) and \(\mathrm{B}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(-5,1)\) is
\(\frac{y-4}{4-1}=\frac{x-1}{1+5}\)
\(\frac{y-4}{3}=\frac{x-1}{6}\)
\(\Rightarrow 2(y-4)=x-1\)
\(\Rightarrow x-1-2 y+8=0\)
\(x-2 y+7=0 \tag{i}\)
Also given equation of line is
\(4 x+3 y-5=0 \tag{ii}\)
These two line intersect each other.
So, solving (i) and (ii) we get \(x=-1, y=3\)
MHT CET-2020
Straight Line
88644
The equation of the line parallel to the line \(3 x-\) \(4 y+2=0\) and passing through \((-2,3)\) is
1 \(3 x-4 y+18=0\)
2 \(3 x-4 y-18=0\)
3 \(3 x+4 y+18=0\)
4 \(3 x+4 y-18=0\)
Explanation:
(A) : Given the equation of the line, \(3 x-4 y+2=0\) \(4 y=3 x+2\)
\(y=\left(\frac{3}{4}\right) x+\frac{1}{2}\)
\(\therefore\) The slope of given line is \(\left(\frac{3}{4}\right)\)
slope of given line \(=\) slope of parallel line \(=\frac{3}{4}\)
Equation of line passes through \((-2,3)\) with slope \(\left(\frac{3}{4}\right)\) is
\((y-3)=\frac{3}{4}(x+2)\)
\(4 y-12=3 x+6\)
\(3 x-4 y+18=0\)
Karnataka CET-2018
Straight Line
88645
If lines represented by \(x+3 y-6=0,2 x+y-4\) \(=0\) and \(k x-3 y+1=0\) are concurrent, then the value of \(k\) is
88646
If the lines \((p-q) x^{2}+2(p+q) x y+(q-p) y^{2}=\) 0 are mutually perpendicular, then
1 \(p=q\)
2 \(\mathrm{p}=0\)
3 \(q=0\)
4 p and q may have any value
Explanation:
(D) : Given, \((p-q) x^{2}+2(p+q) x y+(q-p) y^{2}=0\)
\(\because\) these lines are perpendicular to each other.
\(\therefore\) Coefficient of \(\mathrm{x}^{2}+\) Coefficient of \(\mathrm{y}^{2}=0\)
\(\Rightarrow \mathrm{p}-\mathrm{q}+\mathrm{q}-\mathrm{p}=0\)
\(\therefore \mathrm{p}\) and \(\mathrm{q}\) may have any value.
