Question

A straight line $L$ is drawn through the point $A(2,1)$ such that its point of intersection with the straight line $x + y = 9$ is at a distance of $3\sqrt 2$ from $A$. Find the angle which the line $L$ makes with the positive direction of the $X$- axis.

Answer 1

Here to find the angle we will use the slope formula, initially we will assume the point of intersection and using the given distance we will find the assumed points and find slope with it.

Formula used: We know that the distance between two points $A({x_1},{y_1})$ and $B({x_2},{y_2})$ is $\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}}$
We know that the slope can be written as $\tan \theta$.

Complete step-by-step answer:
It is given that: a straight line $L$ is drawn through the point $A(2,1)$ such that its point of intersection with the straight line $x + y = 9$. The distance of $A$ from the point of intersection is $3\sqrt 2$
Now we have to find the measurement of the angle that the line $L$ makes with the positive direction of the $X$- axis.
Let us assume the point of intersection be $P(h,k)$.
Since, the point $P(h,k)$ lies on the straight line $x + y = 9$, it satisfies the equation of the straight line.
So let us substitute $x = h,y = k$ in the equation of the straight line $x + y = 9$ we get,
$h + k = 9$
Let us find k in terms of h, we get,
$k = 9 - h$
So, the point of intersection becomes $P(h,9 - h)$
We know that the distance between two points $A({x_1},{y_1})$ and $B({x_2},{y_2})$ is $\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2}}$
So, the distance between $P(h,9 - h)$ and $A(2,1)$ we get,
$\sqrt {{{(h - 2)}^2} + {{(9 - h - 1)}^2}}$
Since the distance between the two points is given as $3\sqrt 2$ we get,
$\sqrt {{{(h - 2)}^2} + {{(9 - h - 1)}^2}} = 3\sqrt 2$
Let us simplify the terms inside the square root we get,
$\sqrt {{{(h - 2)}^2} + {{(8 - h)}^2}} = 3\sqrt 2$
By squaring both sides, we get,
${(h - 2)^2} + {(8 - h)^2} = 18$
Let us simplify the above equation we get,
$2{h^2} - 20h + 50 = 0$
Let us divide both sides of the above equation by $2$ we get,
${h^2} - 10h + 25 = 0$
Let us solve using algebraic identity we have,
${(h - 5)^2} = 0$ gives, $h = 5$
Substitute the value of h in $k = 9 - h$ we get, $k = 4$
Hence, the point of intersection is $P(5,4)$.
We have found that the straight line $L$ passes through the points $P(5,4)$ and $A(2,1)$.
Slope of the line is given by the formula $\dfrac{{{y_1} - {y_2}}}{{{x_1} - {x_2}}}$ here ${x_1} = 5,{y_1} = 4\& {x_2} = 2,{y_2} = 1$
So slope of the line is $\dfrac{{4 - 1}}{{5 - 2}} = \dfrac{3}{3} = 1$
Let us take the angle which the line $L$ makes with the positive direction of the $X$- axis as $\theta$.
We know that the slope can be written as $\tan \theta$.
As per the problem,
$\tan \theta = 1$ gives, $\theta = \dfrac{\pi }{4}$.
Hence, the positive direction of the $X$- axis is$\dfrac{\pi }{4}$.

Note: The distance between two points will be always positive hence we will not assume negative values in the given solution.