Study guides › 0580 › Coordinate geometry Maths4U Original · Learn, check, explain
Perpendicular straight lines — 0580
Use a negative reciprocal and then impose the stated point.
Extended · Syllabus E3.7
1. Understand 2. See examples 3. Check answers 4. Explain your working Notes and key methods Two nonvertical perpendicular lines have gradients whose product is −1. Thus a nonzero gradient M has perpendicular gradient \(\displaystyle - \frac{1}{M}\). Negating a gradient alone is generally insufficient.
Find the new intercept from the point the new line must contain. Check both the point substitution and the gradient product. Horizontal and vertical lines form a separate perpendicular pair.
Check the conditions Use the stated units and precision. Show the method as well as the final answer. Three worked examples Example 1 · Perpendicular straight lines Line l is \(y=5 - 4 x\). A perpendicular line h passes through \(\displaystyle P=(0,7)\).
Find the gradient of h. Find the y-intercept of h. Find the x-intercept of h. Find y on h when \(\displaystyle x=1\). Show the reasoning (a) Use the negative reciprocal of the nonzero original gradient.
\(\displaystyle M_{h}= - \frac{1}{M_{l}}= \frac{1}{4}\)
Answer: \(\displaystyle \frac{1}{4}\).
(b) Substitute P into the perpendicular-line rule.
\(c_h=7\)
Answer: \(7\).
(c) Set \(\displaystyle y=0\) in the new line equation.
\(\displaystyle 0=\frac{x}{4} + 7\) gives \(\displaystyle x=\)\(-28\).
Answer: \(-28\).
(d) Evaluate the complete new line equation.
One horizontal step from P changes y by \(\displaystyle \frac{1}{4}\), giving \(\displaystyle \frac{29}{4}\).
Answer: \(\displaystyle \frac{29}{4}\).
Example 2 · Perpendicular straight lines Line l is \(\displaystyle y=\frac{x}{2} + 6\). A perpendicular line h passes through \(\displaystyle P=(1,1)\).
Find the gradient of h. Find the y-intercept of h. Find the x-intercept of h. Find y on h when \(\displaystyle x=2\). Show the reasoning (a) Use the negative reciprocal of the nonzero original gradient.
\(\displaystyle M_{h}= - \frac{1}{M_{l}}= - 2\)
Answer: \(-2\).
(b) Substitute P into the perpendicular-line rule.
\(c_h=3\)
Answer: \(3\).
(c) Set \(\displaystyle y=0\) in the new line equation.
\(0=3 - 2 x\) gives \(\displaystyle x=\)\(\displaystyle \frac{3}{2}\).
Answer: \(\displaystyle \frac{3}{2}\).
(d) Evaluate the complete new line equation.
One horizontal step from P changes y by \(-2\), giving \(-1\).
Answer: \(-1\).
Example 3 · Perpendicular straight lines Line l is \(\displaystyle y=7 - \frac{2 x}{3}\). A perpendicular line h passes through \(\displaystyle P=(2,2)\).
Find the gradient of h. Find the y-intercept of h. Find the x-intercept of h. Find y on h when \(\displaystyle x=3\). Show the reasoning (a) Use the negative reciprocal of the nonzero original gradient.
\(\displaystyle M_{h}= - \frac{1}{M_{l}}= \frac{3}{2}\)
Answer: \(\displaystyle \frac{3}{2}\).
(b) Substitute P into the perpendicular-line rule.
\(c_h=-1\)
Answer: \(-1\).
(c) Set \(\displaystyle y=0\) in the new line equation.
\(\displaystyle 0=\frac{3 x}{2} - 1\) gives \(\displaystyle x=\)\(\displaystyle \frac{2}{3}\).
Answer: \(\displaystyle \frac{2}{3}\).
(d) Evaluate the complete new line equation.
One horizontal step from P changes y by \(\displaystyle \frac{3}{2}\), giving \(\displaystyle \frac{7}{2}\).
Answer: \(\displaystyle \frac{7}{2}\).
Common mistakes Keep intermediate values exact until the final step. Check the answer against the original conditions. Practise with answer checking Original exam-style problems Write complete solutions before opening the worked solution. The marking guidance rewards reasoning as well as answers.
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