Question

How to answer this using coordinate geometry- section formula ?

Answer 1

`P (6,3)`

Explanation:

.

`M (10,13)`, `N (4,-2)`, `P (x,y)`

Using the distance formula:

`MN=sqrt((4-10)^2+(-2-13)^2)=sqrt(36+225)=sqrt261`

`MP=sqrt((x-10)^2+(y-13)^2)`

`PN=sqrt((4-x)^2+(-2-y)^2)`

`MP=2/3MN`, and `PN=1/3MN`

`(MN)^2=261`

`(MP)^2=4/9(MN)^2=4/6(261)=116`

`(PN)^2=1/9(MN)^2=1/9(261)=29`

`(MP)^2=x^2-20x+100+y^2-26y+169=116`

`(PN)^2=x^2-8x+16+y^2+4y+4=29`

`1)` `x^2+y^2-20x-26y=-153`

`2)` `x^2+y^2-8x+4y=9`

Subtracting equation `2` from equation `1`, we get:

`-12x-30y=-162`. Dividing it by `6` we get:

`2x+5y=27`

`y=(27-2x)/5`

We substitute this for `y` in equation `2`:

`x^2+((27-2x)/5)^2-8x+4((27-2x)/5)=9`

`x^2+(729-108x+4x^2)/25-8x+(108-8x)/5-9=0`

We multiply the equation by `25` to get rid of fractions:

`25x^2+729-108x+4x^2-200x+540-40x-225=0`

Combining like terms, we get:

`29x^2-348x+1044=0`

Dividing it by `29`, we get:

`x^2-12x+36=0`

`(x-6)^2=0`

`x=6`

`y=(27-2x)/5=(27-2(6))/5=(27-12)/5=15/5=3`

`P (6,3)`

Answer 2

`P=(6,3)`

Explanation:

`"given a line MN being divided by a point P"`
`"in the ratio m : n"`

`"if "M=(x_1,y_1)" and "N=(x_2,y_2)" then P "`

`"using the "color(blue)"section formula"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(P=((nx_1+mx_2)/(m+n)),((ny_1+my_2)/(m+n)))color(white)(2/2)|)))`

`2MN=3MPrArrMP=2/3MN" and "PN=1/3MN`

`(x_1,y_1)=(10,13)" and "(x_2,y_2)=(4,-2)`

`"and the ratio "m:n=2:1`

`rArrP=((10+(2xx4))/(2+1),(13+(2xx-2))/(2+1))`

`color(white)(rArrP)=(18/3,9/3)=(6,3)`