Pregunta
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Question 4
The gradient of the line joining the two points and is .
a) Write down the gradient of the line which is parallel to .
b) Write down the gradient of the line which is perpendicular to .
c) Find the equation of the line parallel to and passing through the point .
Give your final answer in standard form .
d) Find the equation of the line perpendicular to and passing through the point
. Give your final answer in standard form .

Ask by Harris Harmon. in Dominica
Mar 31,2025

Solución de inteligencia artificial de Upstudy

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(a) The gradient of the line parallel to is .
(b) The gradient of the line perpendicular to is .
© The equation of the line parallel to passing through is .
(d) The equation of the line perpendicular to passing through is .

Solución

(a) The gradient of the line parallel to is the same as the gradient of , so it is
(b) The gradient of the line perpendicular to is the negative reciprocal of , so it is
© To find the equation of the line parallel to that passes through , we start with the point-slope form:
Multiplying both sides by 4 to eliminate the fraction gives:
Rearrange the equation into standard form :
(d) To find the equation of the line perpendicular to that passes through , we use the slope in the point-slope form:
Multiply both sides by 3:
Rearrange into standard form:

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Beyond the Answer

The gradient of the line parallel to will be the same, so it is also . For a line to be perpendicular, its gradient must be the negative reciprocal, which would be . So, we have our gradients: parallel line gradient and perpendicular line gradient .
To find the equation of the line parallel to and passing through the point , we can use the point-slope form of the line equation:
which becomes:
Simplifying leads to:
Now, rearranging in standard form :
For the equation of the line perpendicular to passing through the point , use the same point-slope form:
This simplifies to:
Changing to a fraction gives:
Rearranging gives:
And there you have it! The equations are:
  1. Parallel line:
  2. Perpendicular line:

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