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  • Let C the centroid of the triangle with vertices  Let P be the point of intersection of t

Let C the centroid of the triangle with vertices (3,-1),(1,3)\: and \: (2,4). Let P be the point of intersection of the lines x+3y-1 =0 and 3x-y+1 =0. Then the line passing through the points C and P also passes through the point:
Option: 1 (-9,-7)
Option: 2 (-9,-6)
Option: 3 (7,6)
Option: 4 (9,7)
 

Answers (1)

best_answer

 

 

Centroid -

Centroid   

Centroid  of a triangle is the point of intersection of the medians of the triangle. A centroid divides the median in the ratio 2:1.

Whereas, the median is the line joining the mid-points of the sides and the opposite vertices.

The coordinates of the centroid of a triangle (G) whose vertices are A (x1, y1), B (x2, y2) and C(x3, y3), is given by 

\\\mathrm{\mathbf{\left ( \frac{x_1+x_2+x_3}{3},\;\frac{y_1+y_2+y_3}{3} \right )}}

If D (a1, b1), E (a2, b2) and F (a3, b3) are the mid point of ΔABC, then its centroid is given by

\\\mathrm{\mathbf{\left ( \frac{a_1+a_2+a_3}{3},\;\frac{b_1+b_2+b_3}{3} \right )}} 

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Point of intersection of two lines -

Point of intersection of two lines

Equation of two non-parallel line is 

\\L_1=a_1x+b_1y+c_1=0\\L_2=a_2x+b_2y+c_2=0

If P (x1, y1) is a point of intersection of L1 and L2 , then solving these two equations of the line by cross multiplication

\frac{x_1}{b_1c_2-c_1b_2}=\frac{y_1}{c_1a_2-a_1c_2}=\frac{1}{a_1b_2-b_1a_2}

We get,

\mathbf{\left ( x_1,y_1 \right )=\left ( \frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1},\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1} \right )}

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Equation of Straight Line (Part 2) -

Equation of Straight Line

(c) Two-point form

The equation of a straight line passing through the two given points (x1,y1) and  (x1,y1)is  given by

.

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The centroid of triangle ABC D(2,2)

Point of intersection P \left ( -\frac{1}{5},\frac{2}{5} \right )

equation of line DP is  8x – 11y + 6 = 0

Point  (–9,–6) satisfies the equation

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