12. The line segment AB meets the coordinate axes in points A and B. If point P (3,6) divides
AB in the ratio 2:3, then find the points A and B.?
Answers
Answer:
The points are, A(5,0) & B(0,15)
━━━━━━━━━━━━━
\huge\sf\blue{Given}Given
✭ Line segment AB is divided by point P(3,6) in the ratio 2:3
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\huge\sf\gray{To \:Find}ToFind
◈ The points AB?
━━━━━━━━━━━━━
\huge\sf\purple{Steps}Steps
\large\underline{\underline{\sf Concept}}
Concept
So here we shall use the section formula to find the coordinates of A & B. As we are given that the points A & B meets the axes,i.e point A(x,0) & point B(0,y).You may refer to the attachment to get a clear idea of the section formula
━━━━━━━━━
We shall here use the section formula, that is,
\underline{\boxed{\sf x = \dfrac{m_1 x_2+m_2x_1}{m+n}}}
x=
m+n
m
1
x
2
+m
2
x
1
Similarly,
\underline{\boxed{\sf y = \dfrac{m_1y_2+m_2y_1}{m+n}}}
y=
m+n
m
1
y
2
+m
2
y
1
So here we see that,
◕ \sf m_1 \ \& \ m_1 = 2,3m
1
& m
1
=2,3
◕ \sf x_1 \ \& \ x_2 = x,0x
1
& x
2
=x,0
◕ \sf y_1 \ \& \ y_2 = 0,yy
1
& y
2
=0,y
So we shall first try to find the value of x,
➝ \sf x = \dfrac{m_1 x_2+m_2x_1}{m+n}x=
m+n
m
1
x
2
+m
2
x
1
➝ \sf 3 = \dfrac{2(0)+3(x)}{2+3}3=
2+3
2(0)+3(x)
➝ \sf 3 = \dfrac{0+3x}{5}3=
5
0+3x
➝ \sf 3\times 5 = 3x3×5=3x
➝ \sf 15 = 3x15=3x
➝ \sf \dfrac{15}{3} = x
3
15
=x
➝ \sf \red{x=5}x=5
Now it's the time to find the value of y,
➳ \sf y = \dfrac{m_1y_2+m_2y_1}{m+n}y=
m+n
m
1
y
2
+m
2
y
1
➳ \sf 6 = \dfrac{2(y)+0}{5}6=
5
2(y)+0
➳ \sf 6\times 5 = 2y+06×5=2y+0
➳ \sf 30 = 2y30=2y
➳ \sf \dfrac{30}{2} = y
2
30
=y
➳ \sf \orange{y=15}y=15
\sf \bullet\:\: A = (5,0)∙A=(5,0)
\sf \bullet\:\: B = (0,15)∙B=(0,15)
Answer:
A(5,0) and (0,15)
Step-by-step explanation:
let we A as(x,0) and B as(0,y) as the AB meets the coordinate axes in points A and B
p divides AB in ratio 2:3
by section formula (mx2+nx1/m+n)(my2+ny1/m+n)
=>(3,6)=(2(0)+3(x)/3+3)(2(y)+3(0)/2+3)
=>3x/5=3. ,2y/5=6
=>x=15/3,. y=30/2
=>x=5 ,. y=15
A(x,0).=(5,0)
B(0,y)=(0,15)