Question

The points A ( 5, -2), B (2, 7), and C (-10, 4) are in the xy- plane. Find:
a. the equation of the line passing through A and parallel to BC,
b. the equation of the line passing through B, perpendicular to AC,
c. the point of intersection of the lines in (i) and (ii) above.​​

Answer 1

a. Finding equation of line passing through A and parallel to BC.

$\quad$

Since our line is parallel to BC, slope of the line is,

$\longrightarrow\rm{m_a=\dfrac{y_B-y_C}{x_B-x_C}}$

$\longrightarrow\rm{m_a=\dfrac{7-4}{2-(-10)}}$

$\longrightarrow\rm{m_a=\dfrac{1}{4}}$

Since our line passes through A, the equation will be,

$\longrightarrow\rm{y-y_A=m_a(x-x_A)}$

$\longrightarrow\rm{y-(-2)=\dfrac{1}{4}(x-5)}$

$\longrightarrow\rm{\underline{\underline{x-4y-13=0}}\quad\quad\dots(1)}$

$\quad$

b. Finding equation of line passing through B and perpendicular to AC.

$\quad$

Since our line is perpendicular to AC, slope of the line is,

$\longrightarrow\rm{m_b=-\dfrac{x_A-x_C}{y_A-y_C}}$

$\longrightarrow\rm{m_b=-\dfrac{5-(-10)}{-2-4}}$

$\longrightarrow\rm{m_b=\dfrac{5}{2}}$

Since our line passes through B, the equation will be,

$\longrightarrow\rm{y-y_B=m_b(x-x_B)}$

$\longrightarrow\rm{y-7=\dfrac{5}{2}(x-2)}$

$\longrightarrow\rm{\underline{\underline{5x-2y+4=0}}\quad\quad\dots(2)}$

$\quad$

c. Finding point of intersection of the above two lines.

$\quad$

From (1),

$\longrightarrow\rm{x=4y+13}$

Putting this value of x in (2),

$\longrightarrow\rm{5(4y+13)-2y+4=0}$

$\longrightarrow\rm{18y+69=0}$

$\longrightarrow\rm{y=-\dfrac{23}{6}}$

Then,

$\longrightarrow\rm{x=4\left(-\dfrac{23}{6}\right)+13}$

$\longrightarrow\rm{x=-\dfrac{7}{3}}$

Hence the intersection point is $\bf{\left(-\dfrac{7}{3},\ -\dfrac{23}{6}\right)}.$