3x-4y-16=0 'the straight line a find the slope of the line b) slope of a line which is perpendicular to the above lineis what?
Answers
Step-by-step explanation:
The foot of the perpendicular is (
25
68
,
25
−49
)
Step-by-step explanation:
Find the coordinates of the foot of perpendicular from the point ( -1, 3) to the line 3x-4y-16=0.
\text{The foot of perpendicular from the point ( -1, 3) to the line 3x-4y-16=0 is given by }The foot of perpendicular from the point ( -1, 3) to the line 3x-4y-16=0 is given by
\boxed{\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-(ax_1+by_1+c)}{a^2+b^2}}
a
x−x
1
=
b
y−y
1
=
a
2
+b
2
−(ax
1
+by
1
+c)
\text{Here, }Here,
(x_1,y_1)=(-1,3)(x
1
,y
1
)=(−1,3)
a=3,\,b=-4,\,c=-16a=3,b=−4,c=−16
\implies\frac{x+1}{3}=\frac{y-3}{-4}=\frac{-(3(-1)+(-4)(3)+(-16))}{3^2+(-4)^2}⟹
3
x+1
=
−4
y−3
=
3
2
+(−4)
2
−(3(−1)+(−4)(3)+(−16))
\implies\frac{x+1}{3}=\frac{y-3}{-4}=\frac{-(-3-12-16)}{9+16}⟹
3
x+1
=
−4
y−3
=
9+16
−(−3−12−16)
\implies\frac{x+1}{3}=\frac{y-3}{-4}=\frac{-(-31)}{25}⟹
3
x+1
=
−4
y−3
=
25
−(−31)
\implies\frac{x+1}{3}=\frac{y-3}{-4}=\frac{31}{25}⟹
3
x+1
=
−4
y−3
=
25
31
\text{Now, }Now,
\frac{x+1}{3}=\frac{31}{25}
3
x+1
=
25
31
x+1=\frac{93}{25}x+1=
25
93
x=\frac{93-25}{25}x=
25
93−25
\implies\:x=\frac{68}{25}⟹x=
25
68
\frac{y-3}{-4}=\frac{31}{25}
−4
y−3
=
25
31
y-3=\frac{-124}{25}y−3=
25
−124
y=\frac{-124+75}{25}y=
25
−124+75
\implies\:y=\frac{-49}{25}⟹y=
25
−49
\therefore\text{The foot of the perpendicular is }(\frac{68}{25},\frac{-49}{25})∴The foot of the perpendicular is (
25
68
,
25
−49
)
Step-by-step explanation:
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m 1 =
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m 1 = a+1
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m 1 = a+1b−3
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m 1 = a+1b−3
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m 1 = a+1b−3 Slope of the line 3x−4y−16=0 or y=43
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m 1 = a+1b−3 Slope of the line 3x−4y−16=0 or y=43
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m 1 = a+1b−3 Slope of the line 3x−4y−16=0 or y=43 x−4,m
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m 1 = a+1b−3 Slope of the line 3x−4y−16=0 or y=43 x−4,m 2 = 43
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m 1 = a+1b−3 Slope of the line 3x−4y−16=0 or y=43 x−4,m 2 = 43Since these two lines are perpendicular,m1m2
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m 1 = a+1b−3 Slope of the line 3x−4y−16=0 or y=43 x−4,m 2 = 43Since these two lines are perpendicular,m1m2
Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0Slope of the line joining (−1,3) and (a,b), is m 1 = a+1b−3 Slope of the line 3x−4y−16=0 or y=43 x−4,m 2 = 43Since these two lines are perpendicular,m1m2 =−1
∴{a+1b−3 }⋅{ 43}=−1
}=−1⇒ 4a+4
3b−9
3b−9
3b−9 =−1
3b−9 =−1⇒3b−9=4a−4
3b−9 =−1⇒3b−9=4a−4⇒4a+3b=5...(1)
3b−9 =−1⇒3b−9=4a−4⇒4a+3b=5...(1)Point(a,b) lies on line 3x−4y=16
3b−9 =−1⇒3b−9=4a−4⇒4a+3b=5...(1)Point(a,b) lies on line 3x−4y=16∴3a−4b=16....(2)
3b−9 =−1⇒3b−9=4a−4⇒4a+3b=5...(1)Point(a,b) lies on line 3x−4y=16∴3a−4b=16....(2)On solving equation (1) and (2) we obtain
a= 2568 and b=− 2549
a= 2568 and b=− 2549
a= 2568 and b=− 2549
a= 2568 and b=− 2549 Thus, the required coordinates of the foot of the perpendicular are { 2568 , 2549 }..