haloo mauu buat quiss nih !!
jawabannya jangaan asal asalan yaa !!
kalian buat 10 soal tentang koordinat katersius berserta cara penyelesaian nya !! jangan liat yang di buku paket,ayoo kalian buat sendiri eitss tapii kalo mungkin ada di ulangan harian kalian,ambil dari situ bowlee deh !! intinya jangan asal yaa harus dengan cara penyelesaian, terimaakasii
jawabannya jangaan asal asalan yaa !!
kalian buat 10 soal tentang koordinat katersius berserta cara penyelesaian nya !! jangan liat yang di buku paket,ayoo kalian buat sendiri eitss tapii kalo mungkin ada di ulangan harian kalian,ambil dari situ bowlee deh !! intinya jangan asal yaa harus dengan cara penyelesaian, terimaakasii
Jawaban:
Certainly! Here's an explanation for each of the 10 questions:
1. To find the coordinates of the midpoint between two points A(2, 4) and B(8, 10), we use the midpoint formula: (xt, yt) = ((x1 + x2)/2, (y1 + y2)/2). Plugging in the values, we get (xt, yt) = ((2 + 8)/2, (4 + 10)/2) = (5, 7).
2. To find the distance between two points P(3, 2) and Q(7, 8), we use the Euclidean distance formula: d = √((x2 - x1)^2 + (y2 - y1)^2). Plugging in the values, we get d = √((7 - 3)^2 + (8 - 2)^2) = √(16 + 36) = √52 = 2√13.
3. To find the equation of a line passing through point A(2, 3) with a gradient of 4, we use the point-slope form: y - y1 = m(x - x1). Plugging in the values, we get y - 3 = 4(x - 2), which simplifies to y = 4x - 5.
4. To find the gradient of a line passing through points P(1, 5) and Q(3, 9), we use the gradient formula: m = (y2 - y1)/(x2 - x1). Plugging in the values, we get m = (9 - 5)/(3 - 1) = 4/2 = 2.
5. To find the coordinates of the intersection point between the lines y = 2x + 1 and y = -3x + 5, we set the two equations equal to each other and solve for x. Then, we substitute the value of x into one of the equations to find the corresponding y-value. The intersection point is (4/5, 13/5).
6. To find the gradient of a line perpendicular to the line y = 3x - 2, we take the negative reciprocal of the gradient of the given line. In this case, the gradient of the perpendicular line is -1/3.
7. To find the equation of a line passing through point A(4, -1) and perpendicular to the line y = -2x + 3, we use the point-slope form. Plugging in the values, we get y - (-1) = -1/(-2)(x - 4), which simplifies to y + 1 = 1/2(x - 4).
8. To find the equation of a line parallel to the line 2x - 3y = 5 and passing through point P(4, -2), we use the point-slope form. First, we rearrange the given equation to slope-intercept form: y = (2/3)x - 5/3. Since the parallel line has the same slope, the equation becomes y - (-2) = (2/3)(x - 4), which simplifies to y + 2 = (2/3)x - 8/3.
9. To find the equation of a line perpendicular to the line 3x + 4y = 7 and passing through point Q(2, -1), we use the point-slope form. First, we rearrange the given equation to slope-intercept form: y = (-3/4)x + 7/4. Since the perpendicular line has the negative reciprocal slope, the equation becomes y - (-1) = (4/3)(x - 2), which simplifies to y + 1 = (4/3)x - 8/3.
10. To find the equation of a line passing through points A(1, 2) and B(3, 4), we use the point-slope form. First, we find the gradient using the formula: m = (y2 - y1)/(x2 - x1). Plugging in the values, we get m = (4 - 2)/(3 - 1) = 2/2 = 1. Then, we use the point-slope form with point A: y - 2 = 1(x - 1), which simplifies to y - 2 = x - 1.