I have Problem in Analytical Geometry Question: The Portion of a Straight line between the axes is bisected at the point (-3,2) Find its Equation
My Try I used Equation formula $$(y-y_1)=m(x-x_1)$$ but it requires Slope
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Draw a picture. It is clear that the line segment will have to be in the second quadrant. Let the $x$-intercept be $(-a,0)$ and the $y$-intercept be $(0,b)$. Find $a$ and $b$. The rest should be standard. Alternately, you can read off the slope from the picture. – André Nicolas Sep 17 '15 at 05:10
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@RAM: Given two points of a line, how do you find the slope? – Cameron Buie Sep 17 '15 at 05:13
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Let $a$ & $b$ be the x-intercept & y-intercept respectively then the straight line will intersect the axes at the points $(a, 0)$ & $(0, b)$ respectively then the mid-point $(-3, 2)$ of the intersected portion of the line between the coordinate axes is given as $$(-3, 2)\equiv \left(\frac{a+0}{2}, \frac{0+b}{2}\right)$$ $$(-3, 2)\equiv \left(\frac{a}{2}, \frac{b}{2}\right)$$ Hence, by comparing the coordinates we get $$a=2(-3)=-6, b=2(2)=4$$ Then the equation of the line: using intercept formula of the line $$\frac{x}{a}+\frac{y}{b}=1$$ $$\frac{x}{-6}+\frac{y}{4}=1$$
$$\color{red }{2x-3y+12=0}$$
Harish Chandra Rajpoot
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It is the definition of midpoint; draw a picture and it will be clear. – Kevin Sheng Sep 17 '15 at 05:13
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@Kevin: Alright if you know the intersection points $(a, 0)$ & $(0, b)$ with the axes then the mid-point of intersected portion of the line is given as $$\left(\frac{a+0}{2}, \frac{0+b}{2}\right)$$ Does it really need a picture for clarification? – Harish Chandra Rajpoot Sep 17 '15 at 05:19