9. The equation of a line passing through the origin and perpendicular to the line
7x −3y + 4 = 0 is
(A) 7x −3y + 4 = 0
(B) 3x −7y + 4 = 0
(C) 3x + 7y = 0
(D) 7x −3y = 0
Answers
Answer:
(C)
Step-by-step explanation:
Step 1: the slope of line 7x -3y + 4=0 is 7/3. It is found by using y = mx+c. Make y the subject and find it like this.
7x+4 =3y
=> 7/3x +4/3 =y
Therefore, m = 7/3.
Step 2: The line which passes is perpendicular to the give line. Means its slope is found by:
m2 = -1/m1
So, we have m1 as 7/3 put it in the above formula to get slope of line passing through origin. It will be -3/7.
Step 3: Now since it is passing through origin we can use the formula (y - y1 )= m(x - x1).
so here y1 and x1 are 0 and 0 since it is passing through origin. Substitute the values of m(-3/7) and y1 and x1. The overall answer after solving is:
3x + 7y =0
Hope this helps. Thank you.
Answer:
ION MODEL I
Total Marks 30
521-5
1)
If in triangles ABC and EDF,
AB
BC
then they will be similar, when
FD
(a) ZB= ZE (b) ZA ZD Je) ZB= ZD (d) ZA= 2F
2) The perimeters of two similar triangles AABC and APQR are 36 cm and 24 cm
DE
respectively. If PQ = 10 cm, then the length of AB is
(a) 6 (b)
10/6
3
cm
(c) 60 cm (d) 15cm
3) The point of intersection of 3x - y = 4 and x + y = 8 is
(a) (5, 3) (b) (2, 4) (e) (3, 5) (d) (4, 4)
4) The equation of a line passing through the origin and perpendicular to the
line
(a) 7x - 3y + 4 = 0 (b) 3x - 7y + 4 = 0 Je) 3x + 7y=0 (d) 7x - 3y = 0
5) if sin cos , then 2tan2 0 +sin² 0 -1 is equal to
=
(a) 글 (b) 을 (c) 룸 (d) 긍
6) Find the area of the triangle formed by the points (1,-1), (-4, 6) and (-3,-5)
7) The line through the points (-2, a) and (9, 3) has slope. Find the value of a.
from a point on the ground,
5x2=10
ANSWER ALL
Exam Time: 01:00:00 Hrs