Gujarat Board Solutions for Chapter: Straight Lines, Exercise 3: EXERCISE 10.3

Reduce the given equation into normal form. Find it's perpendicular distance from the origin and angle between the perpendicular and the positive $x$-axis.
 $x-y=4$

Find the distance of the point $\left(-1,1\right)$ from the line $12\left(x+6\right)=5\left(y-2\right).$

Find the points on the $x$-axis, whose distances from the line $\frac{x}{3}+\frac{y}{4}=1$ are $4$ units.

Find the distance between the parallel lines $15x+8y-34=0$ and $15x+8y+31=0$.

Find the distance between parallel lines $l\left(x+y\right)+p=0$ and $l\left(x+y\right)-r=0$.

If $p$ and $q$ are the lengths of perpendiculars from the origin to the lines $x\cos \theta -y\sin \theta =k\cos 2\theta$ and $x\sec \theta +y\mathrm{cosec}\theta =k,$ respectively, prove that $p^2+4q^2=k^2$.

In the triangle $ABC$ with vertices $A\left(2,3\right),B\left(4,-1\right)$ and $C\left(1,2\right),$ find the equation and length of altitude from the vertex $A.$

If $p$ is the length of perpendicular from the origin to the line whose intercepts on the axes are $a$ and $b,$ then show that: $\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}$.