MHT CET · Maths · Straight Lines
PS is the median of the triangle with vertices at \(P(2,2), Q(6,-1)\) and \(R(7,3)\), then the intercepts on the co-ordinate axes of the line passing through point \((1,-1)\) and parallel to PS are respectively
- A \(\frac{7}{2}, \frac{-7}{9}\)
- B \(\frac{2}{7}, \frac{9}{7}\)
- C \(\frac{-7}{2}, \frac{-7}{9}\)
- D \(-2,-9\)
Answer & Solution
Correct Answer
(C) \(\frac{-7}{2}, \frac{-7}{9}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{S}=\) midpoint of \(\mathrm{QR}=\left(\frac{6+7}{2}, \frac{-1+3}{2}\right)=\left(\frac{13}{2}, 1\right)\)
\(\therefore \quad\) 'm' of \(\mathrm{PS}=\frac{2-1}{2-\frac{13}{2}}=-\frac{2}{9}\)
\(\therefore \quad\) The required equation is \(y+1=\frac{-2}{9}(x-1)\)
\(\Rightarrow 2 x+9 y+7=0\)
Here, intercept on \(\mathrm{X}\)-axis is \(-\frac{7}{2}\) and intercept
on \(Y\)-axis is \(-\frac{7}{9}\)
\(\therefore \quad\) 'm' of \(\mathrm{PS}=\frac{2-1}{2-\frac{13}{2}}=-\frac{2}{9}\)
\(\therefore \quad\) The required equation is \(y+1=\frac{-2}{9}(x-1)\)
\(\Rightarrow 2 x+9 y+7=0\)
Here, intercept on \(\mathrm{X}\)-axis is \(-\frac{7}{2}\) and intercept
on \(Y\)-axis is \(-\frac{7}{9}\)
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