Question

Find the distance of the point of intersection of the lines 2x − 3y + 5 = 0 and 3x + 4y = 0 from the line 5x − 2y = 0.

Answer 1

GIVEN:

Equations of two straight lines 2x − 3y + 5 = 0 and 3x + 4y = 0

To find :

The distance of the point of intersection of these lines  from the third line given: 5x − 2y = 0.

The given lines are :

2x-3y=-5

3x+4y=0

Here

$D=\begin{vmatrix} 2 & -3\\ 3 & 4\end{vmatrix} = 8+9=17$

$D_x=\begin{vmatrix} -5 & -3\\0 & 4\end{vmatrix} = -20-0=-20$

$D_y=\begin{vmatrix} 2 & -5\\ 3 & 0\end{vmatrix} = 0-(-15)=-15$

Thus

$x =\frac{D_x}{D}=\frac{-20}{17}\\\\and\\\\y=\frac{D_y}{D}=\frac{-15}{17}$

So the point of intersection is $(\frac{-20}{17} , \frac{15}{17} )$

Now we know that the distance of a point (p,q)

from line ax+by+c=0 is given by

$d=\frac{pa+qb+c} { \sqrt{a^2+b^2}}$

Putting all the value we get

$d =\frac{5 \times \frac {-20}{17}- 2\times \frac{15}{17}}{\sqrt{5^{2}+2^2 }}$

$= - \frac{ 130}{17\sqrt{29} }$

Answer:

Thus the distance of line 5x-2y = 0 from intersection point is :

$d= \frac{ 130}{17\sqrt{29} } \, unit$

Answer 2

$\large\underline\purple{\bold{Solution :- }}$

Let us first find the point of intersection of the lines

$\sf \: 2x - 3y + 5 = 0 - - - (1)$

and

$\sf \: 3x + 4y = 0 - - - (2)$

Now, using method of eliminations,

Multiply (1) by 3 and (2) by 2, and subtract, we get

$\rm :\implies\:6x - 9y + 15 - 6x - 8y = 0$

$\rm :\implies\: - 17y = - 15$

$\rm :\implies\: \boxed{ \pink{ \bf \: y \: = \tt \: \dfrac{15}{17} }}$

On substituting the value of y in equation (1), we get

$\rm :\implies\:3x + 4 \times \dfrac{15}{17} = 0$

$\rm :\implies\:3x = - \dfrac{60}{17}$

$\rm :\implies\: \boxed{ \pink{ \bf \: x \: = \tt \: - \dfrac{20}{17} }}$

$\rm :\implies\: \boxed{ \pink{ \bf \: Hence, \: the \: point \: of \: intersectio \: is ( - \dfrac{20}{17} , \dfrac{15}{17} ) \: }}$

Now,

Let us consider a line ax + by + c = 0 and let (p, q) be any point in the plane, then distance (d) between line and point is given by

$\rm :\implies\: \boxed{ \green{ \bf \: d\: = \tt \: \dfrac{ |ap \: + \: bq \: + \: c| }{ \sqrt{ {a}^{2} + {b}^{2} } } }}$

So, now

Distance of the line 5x - 2y = 0 from the point of intersection is given by

$\rm :\implies\:d = \dfrac{ |5 \times \dfrac{( - 20)}{17} - 2 \times \dfrac{15}{17} | }{ \sqrt{ {5}^{2} + {2}^{2} } }$

$\rm :\implies\:d \: = \: \dfrac{ | - \dfrac{100}{17} - \dfrac{30}{17} | }{ \sqrt{25 + 4} }$

$\rm :\implies\: \boxed{ \pink{ \bf \: d \: = \tt \:\dfrac{130}{17 \sqrt{29} \: } \: units }}$