Eric Nicholas K's answer to Keith6156's Secondary 4 A Maths Singapore question.
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An idea.
You need to find
- the gradient of our wanted line (which is found by finding the gradient of the given other line 3x + 5y = 13 and using the fact that the product of gradients of two perpendicular lines is -1), which is 5/3, and
- a point in this line (given by the intersection of two lines), which is (10/11, 3/11), then
- form the equation using y = mx + c.
You need to find
- the gradient of our wanted line (which is found by finding the gradient of the given other line 3x + 5y = 13 and using the fact that the product of gradients of two perpendicular lines is -1), which is 5/3, and
- a point in this line (given by the intersection of two lines), which is (10/11, 3/11), then
- form the equation using y = mx + c.
Date Posted:
4 years ago
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