Jamia Millia Islamia Straight Line Previous Year Questions (PYQs) – Page 1 of 2
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Coordinates of midpoint of line joining two points (16, 4) and (36, 6) are
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Solution
Midpoint = ((16+36)/2, (4+6)/2) = (26, 5).
The equation of straight line passing through (3, 2) and perpendicular to $y = x$ is
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Solution
Slope of $y = x$ is 1 → perpendicular slope = −1.
Equation: $y - 2 = -1(x - 3) \Rightarrow x + y = 5$.
Specifying a straight line, how many geometrical parameters should be known?
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Solution
A line is determined by two parameters — slope and intercept (or two points).
A point equidistant from lines $4x + 3y + 10 = 0$, $5x - 12y + 26 = 0$, and $7x + 24y - 50 = 0$ is
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Solution
Solving equal distance condition gives point (1,1).
The coordinates of the foot of perpendicular from the point $(2,3)$ on the line $y = 3x + 4$ is given by:
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Solution
Given line: $y = 3x + 4 \Rightarrow 3x - y + 4 = 0$
For point $(x_1, y_1) = (2, 3)$,
Foot of perpendicular $(x, y)$ is given by:
$x = \dfrac{b(bx_1 - ay_1) - ac}{a^2 + b^2}$ and $y = \dfrac{a(-bx_1 + ay_1) - bc}{a^2 + b^2}$
Here $a=3, b=-1, c=4$.
So,
$x = \dfrac{(-1)((-1)(2) - 3(3)) - (3)(4)}{3^2 + (-1)^2} = \dfrac{1(-11) - 12}{10} = \dfrac{37}{10}$
$y = \dfrac{3(-(-1)(2) + 3(3)) - (-1)(4)}{10} = \dfrac{1}{10}$
$\boxed{\text{Answer: (A) }\left(\dfrac{37}{10}, \dfrac{1}{10}\right)}$
Equations of diagonals of the square formed by the lines $x = 0,\ y = 0,\ x = 1$ and $y = 1$ are:
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Solution
Vertices of square: $(0,0), (1,0), (0,1), (1,1)$
Diagonals: $y = x$ and $y + x = 1$
$\boxed{\text{Answer: (A) }y=x,\ y+x=1}$
A point $P$ on the $y$-axis is equidistant from the points $A(-5,4)$ and $B(3,-2)$.
Its coordinate is
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Solution
Let $P(0, y)$.
$\sqrt{(0+5)^2+(y-4)^2} = \sqrt{(0-3)^2+(y+2)^2}$
$\Rightarrow 25 + y^2 - 8y + 16 = 9 + y^2 + 4y + 4$
$\Rightarrow 12y = 28 \Rightarrow y = \tfrac{7}{3}.$
What is the value of $x$ for which the points $(x,-1)$, $(2,1)$ and $(4,5)$ are collinear?
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Solution
For collinearity, slopes must be equal:
$\dfrac{1-(-1)}{2-x} = \dfrac{5-1}{4-2} \Rightarrow \dfrac{2}{2-x} = 2
\Rightarrow 1 = 2 - x \Rightarrow x = 1.$
For which value of $k$, the line $(k-3)x-(4-k^2)y+k^2-7k+6=0$ will be parallel to the $x$-axis?
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Solution
A line parallel to the $x$-axis has no $x$-term ⇒ coefficient of $x$ must be $0$:
$k-3=0 \Rightarrow k=3$.
Also need coefficient of $y \ne 0$: $-(4-k^2)\ne 0 \Rightarrow k\ne \pm 2$ (satisfied).
If the point $P(x,y)$ is equidistant from $A(a+b,\,b-a)$ and $B(a-b,\,a+b)$, then …
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Solution
Set distances equal and square:
$(x-a-b)^{2}+(y-b+a)^{2}=(x-a+b)^{2}+(y-a-b)^{2}$.
Simplify ⇒ $-4b(x-a)+4a(y-b)=0 \Rightarrow bx=ay$.
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