The first equation we were given was 2x + 3y = 26 for L1, and we needed to rearrange that in the first part of the question to find the gradient of L2 (perpendicular to L1). I rearranged it to get y = -2/3x +26/3, which was correct.
We could then work out the equation of L2, which was y=2/3x, which was correct.
However, for part B (this one in the image), you need to substitute them into each other to find where they intersect. ExamSolutions and MathsGenie solutions did it via using L1 in form ax+by+c and L2 in the form y=mx+c, replacing b with mx+c.
However, I tried doing it by using L1 in the form y=-2/3x +26/3 and L2 in form y=2/3x, made them equal to each other and tried it that way, so I had:
2/3x = -2/3x + 26/3
Are you not able to do it the way I tried? I tried it and got a horrible number when solving for x.
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C1 Geometry, should this way of substitution also work? watch
- Thread Starter
- 29-01-2017 13:21
- 29-01-2017 13:29
You have the equation of the second line wrong.
Your method is perfectly valid as actually is exactly he same method as the guy on the video - however you get the wrong answer because your equation of the second line is not right.
- 29-01-2017 13:32
You have to reciprocate the gradient for the perpendicular line