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Question: The Point P is the foot of perpendicular from A(-5,7) to the line \(2x - 3y + 18 = 0\). (i)Determi...

The Point P is the foot of perpendicular from A(-5,7) to the line 2x−3y+18=02x - 3y + 18 = 0.
(i)Determine the equation of the line.
(ii)The coordinate of point P.

Explanation

Solution

To find the equation of the required line, we have to calculate the foot of perpendicular. The property of being perpendicular is the relationship between two lines which meet at a right angle. For the foot of perpendicular coordinates, we have to calculate the slope of line PA. So, firstly we try to find the slope of line PA.

Complete step-by-step answer:
Equation of Line BC =2x−3y+18=0 = 2x - 3y + 18 = 0
⇒2x−3y=−18 ⇒−3y=−2x−18 ⇒y=23x+183 ⇒y=23x+6  \Rightarrow 2x - 3y = - 18 \\\ \Rightarrow - 3y = - 2x - 18 \\\ \Rightarrow y = \dfrac{2}{3}x + \dfrac{{18}}{3} \\\ \Rightarrow y = \dfrac{2}{3}x + 6 \\\
The slope of line BC =23=m1 = \dfrac{2}{3} = {m_1}
Let P(a,b) be the foot of the perpendicular from A(-5,7)
Let m2{m_2} be the slope of the line PA.
Since BC and PA are perpendicular
∴m1m2=−1\therefore {m_1}{m_2} = - 1
⇒23×m2=−1 ⇒m2=−123 ⇒m2=−32  \Rightarrow \dfrac{2}{3} \times {m_2} = - 1 \\\ \Rightarrow {m_2} = \dfrac{{ - 1}}{{\dfrac{2}{3}}} \\\ \Rightarrow {m_2} = \dfrac{{ - 3}}{2} \\\
So Slope of line PA is -3/2.
Now, P(a, b) and A(-5, 7) are coordinates of PA and -3/2 is the slope of PA.
∴ Equation of PA
⇒y−y1=m(x−x1) ⇒(7−b)=−32(−5−a) ⇒2(7−b)=−3(−5−a) ⇒14−2b=15+3a ⇒30+2b=14−15 ⇒3a+2b=−1....................(i)  \Rightarrow y - {y_1} = m(x - {x_1}) \\\ \Rightarrow (7 - b) = \dfrac{{ - 3}}{2}( - 5 - a) \\\ \Rightarrow 2(7 - b) = - 3( - 5 - a) \\\ \Rightarrow 14 - 2b = 15 + 3a \\\ \Rightarrow 30 + 2b = 14 - 15 \\\ \Rightarrow 3a + 2b = - 1....................(i) \\\
P(a, b) is also a point of BC . so it will satisfy BC also
⇒2a−3b+18=0 ⇒2a−3b=−18...................(ii)  \Rightarrow 2a - 3b + 18 = 0 \\\ \Rightarrow 2a - 3b = - 18...................(ii) \\\
Solving (i) and (ii)
3a+2b=−1]×2 2a−3b=−18]×3  3a + 2b = - 1] \times 2 \\\ 2a - 3b = - 18] \times 3 \\\
6a+4b=−2 6a−9b=−54‾ 13b=52 b=5213=4 ⇒b=4  6a + 4b = - 2 \\\ \underline {6a - 9b = - 54} \\\ 13b = 52 \\\ b = \dfrac{{52}}{{13}} = 4 \\\ \Rightarrow b = 4 \\\
Putting the value of b in (i)
⇒3a+2×4=−1 ⇒3a+8=−1 ⇒3a=−1−8 ⇒3a=−9 ⇒a=−3  \Rightarrow 3a + 2 \times 4 = - 1 \\\ \Rightarrow 3a + 8 = - 1 \\\ \Rightarrow 3a = - 1 - 8 \\\ \Rightarrow 3a = - 9 \\\ \Rightarrow a = - 3 \\\
So, coordinate of the foot of perpendicular is P|(-3,4)
Equation of line PA is
⇒y−y1=m(x−x1) ⇒y−4=−32(x+3) ⇒2(y−4)=−3(x+3) ⇒2y−8=−3x−9 ⇒3x+2y+1=0  \Rightarrow y - {y_1} = m(x - {x_1}) \\\ \Rightarrow y - 4 = \dfrac{{ - 3}}{2}(x + 3) \\\ \Rightarrow 2(y - 4) = - 3(x + 3) \\\ \Rightarrow 2y - 8 = - 3x - 9 \\\ \Rightarrow 3x + 2y + 1 = 0 \\\

Note: When two lines are perpendicular then their slope’s product is equal to -1. There are many examples of perpendicular lines in everyday life, including a football field and train tracks. To remember the difference between parallel and perpendicular lines is to look at the l's in parallel. Parallel lines describe the two l's in their name,Identify whether the pairs of lines are parallel, perpendicular, or just intersecting.