Question

A Straight line meets the coordinate axes in A and B. find the equation of the straight line when AB is the divided in the ratio2.3at(-5,2)
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Answer 1

Step-by-step explanation:

Let A(α,0),B(0,β)

Let A(α,0),B(0,β)By section formula,

Let A(α,0),B(0,β)By section formula,−5=

Let A(α,0),B(0,β)By section formula,−5= 5

Let A(α,0),B(0,β)By section formula,−5= 53α

Let A(α,0),B(0,β)By section formula,−5= 53α

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α=

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2=

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 5

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a line

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineα

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx +

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + β

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy =1

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy =1⇒

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy =1⇒ −25

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy =1⇒ −253(x)

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy =1⇒ −253(x)

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy =1⇒ −253(x) +

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy =1⇒ −253(x) + 5

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy =1⇒ −253(x) + 5y

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy =1⇒ −253(x) + 5y

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy =1⇒ −253(x) + 5y =1

Let A(α,0),B(0,β)By section formula,−5= 53α ⇒α= 3−25 and 2= 52β ⇒β=5⇒ By intercept form, equation of a lineαx + βy =1⇒ −253(x) + 5y =1⇒−3x+5y=25