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At (-2,4): 4 = m(-2) + c → c = 4+2m

At (x1,0): 0 = m(x1) + c→ x1 = -c/m

At (0,y1) y1 = m(0) + c → y1 = c

x1+y1 = -4

∴-c/m +c=-4/1

y1=c

c=2m+4

next substitute c value for c into equation: -c/m +c=-4/1

(-2m+4)/m +2m+4=-4

2m+4+2m²+4m=0

since 2m+4=c==y1

then x1+4= 2m²+4m

I still can't get it out any ideas anyone. I guess I went hopelessly wrong

**Sephiroth Valentine**- Replies: 2

Hi, my names Jonathan.I'm studing higher level maths for the leaving Cert in Ireland. I'm finding it really difficult, so i'll apologise in advance for asking too many questions.Nice to meet you all.

**Sephiroth Valentine**- Replies: 6

6(C) A line containg the point(-2,4) has a slope of m which is not equal to zero. The line intercepts the x-axis at (x1,0) and the y axis at (0,y1) If x1 + y1=-4m find the slopes of the two lines that satisfy this condition. Hence find the tangent of the acute angle between these two lines.

I really don't know where to start Tried using the slope formula y2-y1/x2-x1 using the point (-2,4) and (x1,0) and then I used the points (-2,4) and (0,y1) to get another value for m. I put bothe equal to one another but I could'nt get a value to work out. Could any help me please.

**Sephiroth Valentine**- Replies: 2

rstu is a quadrilateral where r = (-1,-5) and s = (13,9)

q(3,-1) lies on the line rs.

(i) The coordinates of u are (-2k,3k)

where k∈R and k>0

The area of the triangle rqu is 28 square units. Find the value of k.

(II) The slope of ts is -3/11 and sr are parallel to tu. Find the coordinates of t.

Would anyone be able to tell me the steps to take to solve these questions or would be able to offer suggestions of how to solve them thanks. Much appreciated

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