TG EAMCET 2 May 2025 Shift 2 Paper

Let slope of line passing through the point (4,−3) be m
And slope of the line joining point (1,1) and (2,3) is =‌

3−1

2−1

=2
∴‌tan‌45∘‌=|‌

m−2

1+2m

|
⇒‌1‌=|‌

m−2

1+2m

|
taking positive sign, we get
1‌=‌

m−2

1+2m

⇒1+2m‌=m−2
⇒‌‌m‌=−3
Taking negative sign, we get
‌−1=‌

m−2

1+2m

‌⇒‌‌m−2=−1−2m⇒3m=1
‌⇒‌‌m=‌

1

3

Since, slope is negative.
∴‌‌m=−3
Equation of line is
‌y+3‌=(−3)(x−4)
⇒‌y+3‌=−3x+12
⇒‌3x+y‌=9
∴x-intercept, when y=0, is 3x=9
⇒‌‌x=3
and y-intercept, when x=0 is y=9
∴‌ Sum of intercepts ‌=3+9=12