Let A (a, b), B (3, 4) and C(–6, –8) respectively denote the centroid circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a + 3, 7b + 5) from the line 2x + 3y – 4 = 0 measured parallel to the line x – 2y – 1 = 0 is.
Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.
Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA
A = (0, 0)
⇒ p(3, 5)
PQ is parallel to x– 2y = 1
Slope of PQ = 1/2
tanθ = 1/2
Q lies on 2x + 3y = 4
⇒
Please subscribe our Youtube channel to unlock this solution.