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Let A(1,0),\; B(6,2) and  C\left ( \frac{3}{2},6 \right ) be the vertices of a triangle ABC. If P is a point inside the triangle ABC such that the triangles APC, APB and BPC  have equal areas, then the length of the line segment PQ, where Q is the point \left ( -\frac{7}{6},-\frac{1}{3} \right ),is  _____.
Option: 1 5
Option: 2 7
Option: 3 9
Option: 4 8
 

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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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Distance between two points -

Distance between two points

Point A (x1, y1) and B (x2, y2) is two point on the plane then distance between them is given by \\\mathrm{\mathbf{|AB|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}}

The distance of a point A (x, y) from the  origin O (0, 0) is given by

\\\mathrm{\mathbf{|OA|=\sqrt{x^2+y^2}}}

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A(1,0)  B(6,2)    C(3/2,6)

Point P is the centroid of triangle ABC 

P(17/6,8/3)

Distance between PQ is 5

Posted by

Kuldeep Maurya

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