AMU MCA Straight Line Previous Year Questions (PYQs)
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Showing 8 questions
A ray of light passing through the point $(1,2)$ is reflected on the $x$-axis at a point $P$ and passes through the point $(5,3)$, then the abscissa of point $P$ is
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Solution
A ray of light passing through the point $(1,2)$ reflects on the $X$-axis at point $A$ and the reflected ray passes through the point $(5,3)$. The coordinate of $A$ is
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Solution
The tangent of the angle between the lines whose intercepts on the axes are respectively $a,-b$ and $b,-a$ is
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Solution
Equations of lines:
$ \frac{x}{a}-\frac{y}{b}=1 $
$ \frac{x}{b}-\frac{y}{a}=1 $
Slopes:
$ m_1=\frac{b}{a},\quad m_2=\frac{a}{b} $
Angle between lines:
$ \tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| $
$ =\left|\frac{\frac{b}{a}-\frac{a}{b}}{1+\frac{b}{a}\cdot\frac{a}{b}}\right| $
$ =\left|\frac{\frac{b^2-a^2}{ab}}{2}\right| $
$ =\left|\frac{b^2-a^2}{2ab}\right| $
$ \therefore \tan\theta=\pm\frac{b^2-a^2}{2ab} $
Let the equation of a straight line passing through point $A(\alpha,\beta,\gamma)$ and having direction ratios $l,m,n$ be
$\dfrac{x-\alpha}{l}=\dfrac{y-\beta}{m}=\dfrac{z-\gamma}{n}=r$
Suppose $P$ is any arbitrary point on this line with coordinates $(\alpha+lr,\beta+mr,\gamma+nr)$. Geometrically, $r$ is
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Solution
Distance $AP = \sqrt{(lr)^2+(mr)^2+(nr)^2}$
$= |r|\sqrt{l^2+m^2+n^2}$
So distance is proportional to $r$.
The two lines $ty = x + t^2$ and $y + tx = 2t + t^3$ intersect at the point lies on the curve whose equation is
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Solution
$ty = x + t^2$ ⇒ $x = ty - t^2$
Substitute:
$y + t(ty - t^2) = 2t + t^3$
$y + t^2y - t^3 = 2t + t^3$
$y(1+t^2) = 2t + 2t^3$
$y = 2t$
Then $x = t(2t) - t^2 = t^2$
So locus: $x = t^2, y = 2t$
$\Rightarrow y^2 = 4x$
A line perpendicular to the line segment joining the points $(1,0)$ and $(2,3)$ divides it in the ratio $1:n$. The equation of the line is
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Solution
Slope of line joining $(1,0)$ and $(2,3)$ = $3$
Perpendicular slope = $-\frac{1}{3}$
Point dividing in ratio $1:n$:
$\left(\frac{n+2}{n+1}, \frac{3n}{n+1}\right)$
Equation:
$y - \frac{3n}{n+1} = -\frac{1}{3}\left(x - \frac{n+2}{n+1}\right)$
Simplify ⇒
$3y + x = \frac{n+11}{n+1}$
The angle between the lines with direction ratios $(4,-3,5)$ and $(3,4,5)$ is
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Solution
$\cos\theta = \frac{4\cdot3 + (-3)\cdot4 + 5\cdot5}{\sqrt{50}\sqrt{50}} = \frac{25}{50} = \frac{1}{2}$
$\theta = \frac{\pi}{3}$
Final Answer: $\boxed{\frac{\pi}{3}}$
Given that the points $P(3,2,-4)$, $Q(5,4,-6)$ and $R(9,8,-10)$ are collinear, the ratio in which $Q$ divides $PR$ externally is
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Solution
$\overrightarrow{PQ}=(2,2,-2)$
$\overrightarrow{QR}=(4,4,-4)$
Ratio = $1:2$
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