JEE MAIN Straight Line Previous Year Questions (PYQs) – Page 7 of 13
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The set of all possible values of $\theta $ in the interval (0, $\pi $) for which the points (1, 2) and (sin $\theta $, cos $\theta $) lie on the same side of the line x + y =1 is :
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Solution
A point on the straight line $3x+5y=15$ which is equidistant from the coordinate axes will lie only in:
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Solution
The equations of two sides $AB$ and $AC$ of a triangle $ABC$ are $4x+y=14$ and $3x-2y=5$, respectively. The point $\left(2,-\frac{4}{3}\right)$ divides the third side $BC$ internally in the ratio $2:1$. The equation of the side $BC$ is
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Let A($-$1, 1), B(3, 4) and C(2, 0) be given three points. A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A1 and A2 be the areas of $\Delta$ABC and $\Delta$PQC respectively, such that A1 = 3A2, then the value of m is equal to :
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In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If ($\alpha$, $\beta$) is the centroid of $\Delta$ABC, then 15($\alpha$ + $\beta$) is equal to :
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Let $A(1,0),; B(2,-1)$ and $C\left(\frac{7}{3},\frac{4}{3}\right)$ be three points.
If equation of bisector of angle $ABC$ is $\alpha x+\beta y=5$, then $\alpha^2+\beta^2$ is:
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Solution
Using internal angle bisector
Slope $=-3$
Equation ⇒ $\alpha=3,; \beta=1$
$\alpha^2+\beta^2=10$
If the shortest distance between the lines
$\dfrac{x-\lambda}{2}=\dfrac{y-4}{3}=\dfrac{z-3}{4}$ and
$\dfrac{x-2}{4}=\dfrac{y-4}{6}=\dfrac{z-7}{8}$ is $\dfrac{13}{\sqrt{29}}$, then a value of $\lambda$ is:1
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Solution
Let the lines
\[
\ell_1:\ \frac{x+5}{3}=\frac{y+4}{1}=\frac{z-\alpha}{-2}
\quad\text{and}\quad
\ell_2:\ 3x+2y+z-2=0\;=\;x-3y+2z-13
\]
be coplanar. If the point $P(a,b,c)$ on $\ell_1$ is nearest to the point $Q(-4,-3,2)$,
then $|a|+|b|+|c|$ is equal to:
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Solution
If the line segment joining the points $(5,2)$ and $(2,a)$ subtends an angle $\dfrac{\pi}{4}$ at the origin, then the absolute value of the product of all possible values of $a$ is:
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Solution
The shortest distance between the lines ${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$ and ${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$, is :
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