1. Indicate in standard form the equation of the line passing through the given point and having the given slope.
E(3, 5), m = no slope
2. Indicate the equation of the given line in standard form.
The line through (2, -1) and parallel to a line with slope of 3/4
3. Indicate the equation of the given line in standard form.
The line containing the hypotenuse of right triangle ABC where A(-5, 5), B(1, 1), and C(3, 4) are the vertices.
4. Indicate in standard form the equation of the line passing through the given points.
P(6, 2), Q(8, -4)
5. Indicate in standard form the equation of the line passing through the given points.
A(4, 1), B(5, 2)
6. Indicate the equation of the given line in standard form.
The line containing the longer diagonal of a quadrilateral whose vertices are A (2, 2), B(-2, -2), C(1, -1), and D(6, 4).
7. Indicate in standard form the equation of the line passing through the given point and having the given slope.
A(5, 5), m = 3
8. Indicate in standard form the equation of the line passing through the given points.
L(5, 0), M(0, 5)
9. Indicate in standard form the equation of the line passing through the given points.
M(0, 6), N(6, 0)
10. Indicate the equation of the given line in standard form.
The line with slope and containing the midpoint of the segment whose endpoints are (2, -3) and (-6, 5).
11. Indicate the equation of the given line in standard form.
The line containing the median of the trapezoid whose vertices are R(-1, 5) , S(1, 8), T(7, -2), and U(2, 0).
12. Indicate in standard form the equation of the line passing through the given points.
R(3, 3), S(-6, -6)
✅ Answers
anyone?
Thanks so much!
Math has never been something i could do.
4
✅ Answers
? Favorite Answer
1)
y=mx+c
m=0
y=c, passes through (3,5)
c=5
line equation is y=5
2)
m=3/4, parallel line have same slope.
y=mx+c
y=3/4x+c
line passes through (2,-1)
c=-5/2
line equation is, 3x-4y=10
3)
AB=sqrt[(-5-1)^2 + (5-1)^2]=sqrt(52)
similarly BC=sqrt(13), CA=sqrt(65)
CA is longest.
Equation of CA is,
(y-3)/(x-4) = (3+5)/(4-5)
I try my best but you must pay attention to each word and line very carefully and even may require you revise this solution 3 times also to clear your misunderstanding. Let the line represented by 8x-5y=0 be AB and the required line through P(-3,-2)and perpendicular to AB be CD.(note that p(-3,-2) is any point on CD. It may be even supposed as C or D).Now ,slope of AB is m1=8/5 (HOW?).let the slope of CD be m. Here AB is PERPENDICULAR to CD. So, m*m1=-1(formula) m*8/5=-1 So, m=-5/8. Now , we are in the position of using point- slope formula for CD(because we now know the SLOPE of CD and a POINT P lying on it). Using the point- slope formula: Y-y1=m(x-x1) Y+2=5/8(x+3) (5/8)x-y=2-15/8 5x-8y=1.This is the required equation.(Which of the form Ax+By=C, you talked about;variables x and y on the left and constant C on the right)
Check your formulae: These are straight forward questions:
1) Since there is slope and point given, then you can use the slope-point form.
y-5 = m * (x-3) ==> y -5 = mx -3m ==> mx – y +5 -3m =0 (standard form).
I think you can solve the following questions with the same line of thought.
First rule: Don’t panic! (They are easy).
1. x=3
2. —->y:3/4(x-2) = ( y+1)
3/4x-3/2-y-1 = -y+3/4x-5/2 = y=.75x-2.5
3. Line from (+3, +4) thru (-5, +5): +X+8Y = +35
4. Line from (+8, -4) thru (+6, +2): +6X+2Y = +40
Line from (+5, +2) thru (+4, +1): +X-Y = +3
7. 3(x-5)=(y-5)
3x-15-y+5 = -y+3x-10 = y=3x-10
8. Line from (+0, +5) thru (+5, +0): +5X+5Y = +25
9. Line from (+6, +0) thru (+0, +6): +6X+6Y = +36
12. Line from (-6, -6) thru (+3, +3): +9X-9Y = +0