JEE MAIN Straight Line Previous Year Questions (PYQs) – Page 1 of 13
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Let the three sides of a triangle be on the lines 4x-7y+10=0, x+y=5 and 7x+4y=15. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x=0, y=0 and x+y=1 is
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Solution
Triangle vertices:
$A\left(-\frac{5}{3},\frac{20}{3}\right),\ B(1,2),\ C\left(\frac{25}{11},\frac{30}{11}\right)$
Slope $AC=-1$ ⇒ altitude from $B$ has slope $1$ ⇒ $y=x+1$
Solving with another altitude gives orthocentre $H_1=(1,2)$
Second triangle $(0,0),(1,0),(0,1)$ ⇒ right angled ⇒ $H_2=(0,0)$
Distance $=\sqrt{(1)^2+(2)^2}=\sqrt{5}$
$\boxed{\sqrt{5}}$
$A\left(-\frac{5}{3},\frac{20}{3}\right),\ B(1,2),\ C\left(\frac{25}{11},\frac{30}{11}\right)$
Slope $AC=-1$ ⇒ altitude from $B$ has slope $1$ ⇒ $y=x+1$
Solving with another altitude gives orthocentre $H_1=(1,2)$
Second triangle $(0,0),(1,0),(0,1)$ ⇒ right angled ⇒ $H_2=(0,0)$
Distance $=\sqrt{(1)^2+(2)^2}=\sqrt{5}$
$\boxed{\sqrt{5}}$
Let the area of $\triangle$ PQR with vertices P(5,4), Q(-2,4) and R(a,b) be 35 square units. If its orthocenter and centroid are $O\!\left(2,\dfrac{14}{5}\right)$ and C(c,d) respectively, then c+2d is equal to:
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Solution
Given $P(5,4),\ Q(-2,4)$ and orthocentre $O\left(2,\frac{14}{5}\right)$
$PQ$ is horizontal ⇒ altitude from $R$ is vertical ⇒ $x=2$
So $a=2$
Area $=\frac{1}{2}\times |PQ|\times \text{height}$
$=\frac{1}{2}\times 7 \times |b-4|=35$
$\Rightarrow |b-4|=10 \Rightarrow b=14 \text{ or } -6$
Using orthocentre condition (slopes), valid point is $R(2,-6)$
Centroid:
$c=\frac{5+(-2)+2}{3}=\frac{5}{3},\quad d=\frac{4+4-6}{3}=\frac{2}{3}$
$c+2d=\frac{5}{3}+\frac{4}{3}=3$
$\boxed{3}$
$PQ$ is horizontal ⇒ altitude from $R$ is vertical ⇒ $x=2$
So $a=2$
Area $=\frac{1}{2}\times |PQ|\times \text{height}$
$=\frac{1}{2}\times 7 \times |b-4|=35$
$\Rightarrow |b-4|=10 \Rightarrow b=14 \text{ or } -6$
Using orthocentre condition (slopes), valid point is $R(2,-6)$
Centroid:
$c=\frac{5+(-2)+2}{3}=\frac{5}{3},\quad d=\frac{4+4-6}{3}=\frac{2}{3}$
$c+2d=\frac{5}{3}+\frac{4}{3}=3$
$\boxed{3}$
A ray of light coming from the point (2, $2\sqrt 3 $) is incident at an angle 30o on the line x = 1 at thepoint A. The ray gets reflected on the line x = 1 and meets x-axis at the point B. Then, the line ABpasses through the point :
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Solution
A straight line $L$ at a distance of $4$ units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of $60^\circ$ with the line $x + y = 0$.
Then an equation of the line $L$ is:
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Solution
Let $a,b,c$ and $d$ be non-zero numbers. If the point of intersection of the
lines $4ax+2ay+c=0$ and $5bx+2by+d=0$ lies in the fourth quadrant and is
equidistant from the two axes then :
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Solution
Let $PS$ be the median of the triangle with vertices $P(2,2)$, $Q(6,-1)$ and
$R(7,3)$. The equation of the line passing through $(1,-1)$ and parallel to
$PS$ is :
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Solution
Let L denote the line in the xy-plane with x and y intercepts as 3 and 1 respectively. Then the image of the point (–1, –4) in this line is :
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Solution
Let $A$ be the point of intersection of the lines $L_{1}:\ \dfrac{x-7}{1}=\dfrac{y-5}{0}=\dfrac{z-3}{-1}$ and $L_{2}:\ \dfrac{x-1}{3}=\dfrac{y+3}{4}=\dfrac{z+7}{5}$. Let $B$ and $C$ be the points on the lines $L_{1}$ and $L_{2}$ respectively such that $AB=AC=\sqrt{15}$. Then the square of the area of the triangle $ABC$ is:
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Solution
Two sides of a parallelogram are along the lines, $x+y=3$ and $x-y+3=0$. If its diagonals intersect at $(2,4)$, then one of its vertices is :
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Solution
The angle between the lines whose direction cosines satisfy the equations
$l+m+n=0$ and $l^{2}=m^{2}+n^{2}$ is :
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Solution
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