In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis. Also, find the co-ordinates of the point of intersection.
Let the point P (x, 0) on x-axis divides the line segment joining A (4, 3) and B (2, -6) in the ratio k: 1.
Using section formula, we have:
`0=(-6k+3)/(k+1)`
`0=-6k+3`
`k=1/2`
Thus, the required ratio is 1: 2.
Also, we have:
`x=(2k+4)/(k+1)`
`x=(2xx1/2+4)/(1/2+1)`
`x=10/3`
Thus, the required co-ordinates of the point of intersection are `(10/3,0)`
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Find the ratio in which the join of (-4, 7) and (3, 0) is divided by the y-axis. Also, find the coordinates of the point of intersection.
`0=(3k-4)/(k+1)`
`3k=4`
`k=4/3` ..............(1)
`y=(0+7)/(k+1)`
`y=7/(4/3+1)` (from eq. 1)
`y=3`
Hence, the required is 4:3 and the required point is S(0, 3)
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Let the point P which is on the x-axis, divide the line segment joining the points A(4, 2) and B(3, -5) in the ratio of m : n. Let the coordinates of P be (x, 0).
By section formula,
y-coordinate = my2+ny1m+n\dfrac{my2 + ny1}{m + n}m+nmy2+ny1
⇒0=m×(−5)+n×2m+n⇒0=−5m+2n⇒5m=2n⇒mn=25⇒m:n=2:5.\Rightarrow 0 = \dfrac{m \times (-5) + n \times 2}{m + n} \\[1em] \Rightarrow 0 = -5m + 2n \\[1em] \Rightarrow 5m = 2n \\[1em] \Rightarrow \dfrac{m}{n} = \dfrac{2}{5} \\[1em] \Rightarrow m : n = 2 : 5.⇒0=m+nm×(−5)+n×2⇒0=−5m+2n⇒5m=2n⇒nm=52⇒m:n=2:5.
Putting value of m : n in section-formula for x-coordinate,
x-coordinate = mx2+nx1m+n\dfrac{mx2 + nx1}{m+ n}m+nmx2+nx1
=2×3+5×42+5=6+207=267.= \dfrac{2 \times 3 + 5 \times 4}{2 + 5} \\[1em] = \dfrac{6 + 20}{7} \\[1em] = \dfrac{26}{7}.=2+52×3+5×4=76+20=726.
Hence, coordinates of P are (267,0)(\dfrac{26}{7}, 0)(726,0) and 2 : 5 is the ratio in which the line joining the points (4, 2) and (3, -5) is divided by the x-axis.