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Home » In what ratio does the x–axis divide the line segment joining the points (– 4, – 6) and (–1, 7)? Find the coordinates of the point of division.
In what ratio does the x–axis divide the line segment joining the points (– 4, – 6) and (–1, 7)? Find the coordinates of the point of division.
CBSE | Class 10 | Coordinate Geometry | Exercise 7.3 | Maths | NCERT Exemplar
In what ratio does the x–axis divide the line segment joining the points (– 4, – 6) and (–1, 7)? Find the coordinates of the point of division.

Solution:

Let 1: k be the ratio in which x-axis divides the line segment joining (–4, –6) and (–1, 7).

Therefore,

x-coordinate is (-1 – 4k) / (k + 1)

y-coordinate is (7 – 6k) / (k + 1)

y coordinate = 0, as P lies on x-axis,

\left( 7\text{ }\text{ }6k \right)\text{ }/\text{ }\left( k\text{ }+\text{ }1 \right)\text{ }=\text{ }0

7\text{ }\text{ }6k\text{ }=\text{ }0

k\text{ }=\text{ }6/7

Now, the value of m1 = 6 and m2 = 7

Using section formula, we get,

x\text{ }=\text{ }({{m}_{1}}{{x}_{2}}~+\text{ }{{m}_{2}}{{x}_{1}})/({{m}_{1}}~+\text{ }{{m}_{2}})

=\text{ }\left( 6\left( -1 \right)\text{ }+\text{ }7\left( -4 \right) \right)/\left( 6+7 \right)

=\text{ }\left( -6-28 \right)/13

=\text{ }-34/13

So,

y\text{ }=\text{ }\left( 6\left( 7 \right)\text{ }+\text{ }7\left( -6 \right) \right)/\left( 6+7 \right)

=\text{ }\left( 42-42 \right)/13

=\text{ }0

As a result, the coordinates of P are (-34/13, 0)

i

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