Co-ordinate Geometry Q

copynpaste

New Member

Jul 5, 2013

Hi, I'm having trouble solving this question and some help would be greatly appreciated.

$23. m = \frac{y-2}{x+1}\\mx+m=y-2\\mx-y+m+2=0\\\\x-$intercept: When y = 0:$\\mx+m+2=0\\x=\frac{2-m}{m}\\\\(\frac{-2-m}{m}, 0)\\y-$intercept: When x = 0:$\\-y+m+2=0\\y = m+2\\(0,m+2)\\\\$

$$a) If (-1,2) is the midpoint of the two intercepts, then the intercepts must be (-2,0) and (0,4).\\ Using the intercept formula:$\\\frac{x}{-2} + \frac{y}{4} = 1\\-2x + y = 4\\2x-y+4 = 0\\\\$b) If (-1,2) divides the intercepts in 2:1, then:$\\x = \frac{nx_{1}+mx_{2}}{m+n}\\-1 = \frac{\frac{-2-m}{m} + 2 \times 0}{2+1}\\-1 = \frac{-2-m}{3m}\\3m = 2+m\\m = 1\\\\\therefore $Equation: $x-y+1+2=0\\x-y+3=0\\$(You can use the y-intercept to check answer)$\\\\$

$$c) If (-1,2) divides the intercepts in -2:5, then similarly:$\\x = \frac{nx_{1}+mx_{2}}{m+n}\\-1 = \frac{5 \times \frac{-2-m}{m} + -2 \times 0}{-2+5}\\-1 = \frac{-10-5m}{3m}\\3m = 10+5m\\2m = -10\\m = -5\\\\\therefore $Equation: $-5x-y-5+2=0\\5x+y+3=0$

Last edited: Jul 6, 2013

Co-ordinate Geometry Q (1 Viewer)

copynpaste

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ymcaec

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