JEE MAIN Straight Line Previous Year Questions (PYQs) – Page 9 of 13
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The number of integral values of m so that the abscissa of point of intersection of lines 3x + 4y = 9 and y = mx + 1 is also an integer, is :
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Solution
The equation of one of the straight lines which passes through the point (1, 3) and makes an angles ${\tan ^{ - 1}}\left( {\sqrt 2 } \right)$ with the straight line, y + 1 = 3${\sqrt 2 }$ x is :
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Solution
Let the vertices $Q$ and $R$ of the triangle $PQR$ lie on the line
$\dfrac{x + 3}{5} = \dfrac{y - 1}{2} = \dfrac{z + 4}{3}$,
$QR = 5$ and the coordinates of the point $P$ be $(0, 2, 3)$.
If the area of the triangle $PQR$ is $\dfrac{m}{n}$, then:
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Solution
Let $B$ and $C$ be the two points on the line $y+x=0$ such that $B$ and $C$ are symmetric with respect to the origin.
Suppose $A$ is a point on $y-2x=2$ such that $\triangle ABC$ is an equilateral triangle.
Then, the area of the $\triangle ABC$ is:
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A variable line $L$ passes through the point $(3,5)$ and intersects the positive coordinate axes at the points $A$ and $B$. The minimum area of the triangle $OAB$, where $O$ is the origin, is:
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The line that is coplanar to the line $\dfrac{x+3}{-3}=\dfrac{y-1}{1}=\dfrac{z-5}{5}$ is:
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Let a triangle be bounded by the lines L1 : 2x + 5y = 10; L2 : $-$4x + 3y = 12 and the line L3, which passes through the point P(2, 3), intersects L2 at A and L1 at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to :
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Solution
Let $A=(a,b)\in L_2\Rightarrow -4a+3b=12.$
Since $AP:PB=1:3$, by section formula $P=\dfrac{B+3A}{4}\Rightarrow B=4P-3A=(8-3a,\;12-3b).$
Because $B\in L_1$, $2(8-3a)+5(12-3b)=10\Rightarrow 2a+5b=22.$
Solve
\[
\begin{cases}
2a+5b=22,\\
-4a+3b=12
\end{cases}
\Rightarrow a=\dfrac{3}{13},\quad b=\dfrac{56}{13}.
\]
Thus
\[
A=\left(\dfrac{3}{13},\dfrac{56}{13}\right),\quad
B=\left(\dfrac{95}{13},-\dfrac{12}{13}\right).
\]
Intersection $C=L_1\cap L_2$:
\[
\begin{cases}
2x+5y=10,\\
-4x+3y=12
\end{cases}
\Rightarrow C=\left(-\dfrac{15}{13},\dfrac{32}{13}\right).
\]
Area
\[
\Delta=\frac12\left|
\begin{vmatrix}
x_A&y_A&1\\
x_B&y_B&1\\
x_C&y_C&1
\end{vmatrix}\right|
=\frac12\left|(B-A)\times(C-A)\right|
=\frac12\left|(92)(-24)-(-68)(-18)\right|
=\frac{1716}{169}
=\boxed{\dfrac{132}{13}}.
\]
Answer: $\boxed{\dfrac{132}{13}}$.
Let $(\alpha,\beta)$ be the centroid of the triangle formed by the lines $15x-y=82$, $6x-5y=-4$ and $9x+4y=17$. Then $\alpha+2\beta$ and $2\alpha-\beta$ are the roots of the equation:
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Solution
A light ray emits from the origin making an angle $30^\circ$ with the positive $x$-axis.
After getting reflected by the line $x+y=1$, if this ray intersects the $x$-axis at $Q$, then the abscissa of $Q$ is:
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Solution
$y=\tan30^\circ,x=\dfrac{x}{\sqrt3}$ hits the mirror $x+y=1$ at
$P\left(\dfrac{\sqrt3}{\sqrt3+1},,\dfrac{1}{\sqrt3+1}\right)$.
The mirror’s normal is along $(1,1)$, so reflecting the unit direction $u=(\cos30^\circ,\sin30^\circ)=\left(\dfrac{\sqrt3}{2},\dfrac12\right)$ about the line gives
$u'=u-2(u\cdot \hat n)\hat n=\left(-\dfrac12,-\dfrac{\sqrt3}{2}\right)$,
i.e. slope $m'=\sqrt3$.
The reflected ray through $P$ is $y-y_0=\sqrt3(x-x_0)$.
Intersecting $y=0$ gives $x=x_0-\dfrac{y_0}{\sqrt3} $
$=\dfrac{\sqrt3}{\sqrt3+1}-\dfrac{1}{\sqrt3(\sqrt3+1)}$
$=\dfrac{2}{3+\sqrt3}$
Slope of a line passing through $P(2, 3)$ and intersecting the line $x + y = 7$ at a distance of $4$ units from $P$, is:
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Solution
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